# NCERT Solutions Class 11 Maths Chapter 9

## NCERT Solutions for Class 11 Mathematics Chapter 9 – Sequences and series

Mathematics is an important aspect of every field. It is majorly used in various disciplines of life. The applications of Mathematics make day-to-day dealings of life easier. Hence, it is included as a core subject in almost every academic curriculum.

You have already studied the various patterns of numbers and alphabets in the lower classes. It is nothing but an arrangement with certain repetitions considered as the part of sequence and series. The main topics covered in the chapter are the introduction to sequence and series, Arithmetic Progression (A.P.), Geometric Progression (G.P.), the relationship between A.M and G.M and sum to ‘n’ terms of special series.

NCERT Solutions for Class 11 Mathematics Chapter 9 has detailed coverage of every topic covered in the NCERT Class 9th Mathematics textbook. Students will be able to understand all the concepts related to sequence and series once you refer to NCERT Solutions. How the chapter is presented in the NCERT Solutions for Class 11 Mathematics Chapter 9 will help students gain interest in the subject and assist them in acquiring related knowledge. NCERT solutions will definitely prove fruitful to students in their class assignments, tests and preparation.

Extramarks has been the perfect learning platform for students to improve their grades. The 360-degree learning pattern aids students in learning every concept precisely and quickly without putting much effort, and they start learning in a fun way. To avail of all the NCERT-related study material, they can visit the Extramarks’ website and they must register themselves now, to begin their preparation without any further delay.

### Key Topics Covered In Chapter 9 Class 11 Mathematics

The arrangement of numbers, objects, and alphabets in a particular pattern are called sequence, whereas placing of numbers, objects, and alphabets one after the other without repetition of the pattern is called series. The chapter sequence and series are a combination of both. To get a particular series and understand its associated sequence, one needs to know certain rules based on it. Students will get a clear understanding of this concept in this chapter.

They would be able to derive a series with the help of a few formulas given in the chapter. This will make their calculations while finding a series easier and save their time. The various patterns covered for calculating series in this chapter are AP and GP. Every bit of this chapter is included in our NCERT Solutions for Class 11 Mathematics Chapter 9, available on the  Extramarks’ website.

NCERT Solutions for Class 11 Mathematics Chapter 9 require students to use their critical thinking ability and apply a wide range of formulas they have learnt.

Introduction

In this chapter, we will learn about the sequence and series. When we put a collection of objects in an orderly manner in a certain pattern using some sets of rules, it is called a sequence and series. The sequence and series have an important application in many human activities. When the sequence follows a pattern, it’s called a progression.

We have already studied the arithmetic progression in our previous classes. In this chapter, we will learn some more concepts about the arithmetic progression, series, geometric progression, the relationship between arithmetic and geometric progression and the sum of ‘n’ terms of the series.

Sequences

In the section of this chapter, we will understand the sequence with the help of some examples listed below:

• 2, 4, 8, 16, 32, …………, 1024
• 3, .3, .33, .333, .3333……and so on.

The different elements in a sequence are called terms, for example (a1, a2, a3, a4,…… an)

There are two types of sequences:

• Infinite sequence

When the sequence contains an infinite number of elements, then it is called an infinite sequence.

• Finite sequence

When the sequence contains a finite number of elements, it’s called a finite sequence.

Fibonacci sequence :

Example-

A1 = A2 = 1,

A3 = A1 + A2

An = An-2 + An-1,

n > 2

Series

In this section, we will learn about the series.

There is an example given below that shows the series:

a1 +  a2 +  a3 +  a4 + …… an + …..

The example of the series is associated with the sequence given above.

The important points you should note about the series are:

• The sequence of the series can be finite or infinite.
• The series is offered in the compact form called sigma notation, denoted by the ‘∑’.

Arithmetic progression (A.P)

In the section of this chapter, we will learn about the arithmetic progression or arithmetic sequence. There is a series given below which follows arithmetic progression:

a1 +  a2 +  a3 +  a4 + …… an + …..

an+1 = an + d

Where a = first term, d = common difference of A.P

There are some properties that verify an A.P. They are as follows:

• If a constant is added to each term of A.P., then the resulting sequence will be in A.P
• If a constant is subtracted from each term of A.P., then the resulting sequence will be in A.P
• If a constant is multiplied by each term of A.P., then the resulting sequence will be in A.P
• If a constant/nonzero is divided fry each term of A.P., then the resulting sequence will be in A.P

The series given below follows arithmetic progression:

a, a + d, a + 2d, a + 3d…a + (n + 1)d

Formulas,

• The last number of A.P,

l = a + (n – 1)d

• The sum of n numbers in A.P,

S= n/2 [2a + (n – 1)d]

Or

S= n/2 [a + l]

Arithmetic mean

If we have a sequence, say a, b & c, then the arithmetic mean is given by b = (a + c) /2

Geometric progressions (G.P)

In this part of the chapter, we will learn about the Geometric progression or Geometric sequence.

There is a series given below which follows arithmetic progression:

a1 +  a2 +  a3 +  a4 + …… an + …..

ak+1 / ak = r (constant), for k 1

Where, a = first term, r = common ratio of G.P

The series given below follows a Geometric progression:

a, ar, ar2, ar3,…..

Formulas,

• General ‘n’ term of a G.P,

a=  arn-1

• The sum of n numbers in G.P,

S= a(1-rn) / (1-r)

Or

S= a(rn-1) / (r-1)

Geometric mean

If we have a sequence, say a, b & c, then the geometric mean is given by b = √a.c

Relationship between A.M and G.M

Let us check A.M & G.M.’s relationship of two given numbers a and b.

A.M = (a + b) /2

G.M = √a.b

Then, A.M – G.M = (a + b) /2 – √a.b = (√a – √b)2 / 2 0.

This is the relationship between A.M G.M

Sum to ‘n’ terms of special series.

There is a special series, and we will find the sum of ‘n’ terms of this series.

• 1 + 2 + 3 + 4 +…….+n (sum of first ‘n’ numbers)

Sn = n(n + 1) / 2

• 12 + 22 + 32 + 42 +…….+n2 (sum of first square ‘n’ numbers)

Sn = n(n + 1)(2n + 1) / 6

• 13 + 23 + 33 + 43 +…….+n3(sum of first cube ‘n’ numbers)

Sn = [n(n + 1)]2 / 4

### NCERT Solutions for Class 11 Mathematics Chapter 9 Exercise & Solutions

The NCERT Solutions for Class 11 Mathematics Chapter 9 is mostly practice-oriented. The more you practice, the better you will get. . Hence, multiple exercises are given in this chapter in the NCERT textbook. You must have all the accurate solutions to these exercises. As a result, we have provided a detailed solution to every question given in the chapter in the NCERT Solutions for Class 11 Mathematics Chapter 9 available on the Extramarks’ website. The solutions will give you a complete analysis of the steps you should follow when solving the problems related to sequence and series. Thus, helping you to develop clear mindsets while solving the problems. .

Click on the  link below  to view exercise-specific questions and solutions for NCERT Solutions for Class 11 Mathematics Chapter 9:

• Chapter 9 Class 11 Mathematics: Exercise 9.1
• Chapter 9 Class 11 Mathematics: Exercise 9.2
• Chapter 9 Class 11 Mathematics: Exercise 9.3
• Chapter 9 Class 11 Mathematics: Exercise 9.4
• Chapter 9 Class 11 Mathematics: Miscellaneous Exercise

Along with Class 11 Mathematics solutions, you can explore NCERT Solutions on our Extramarks’ website for all primary and secondary classes.

• NCERT Solutions Class 1,
• NCERT Solutions Class 2,
• NCERT Solutions Class 3,
• NCERT Solutions Class 4,
• NCERT Solutions Class 5,
• NCERT Solutions Class 6,
• NCERT Solutions Class 7,
• NCERT Solutions Class 8,
• NCERT Solutions Class 9,
• NCERT Solutions Class 10,
• NCERT Solutions Class 11.
• NCERT Solutions Class 12.

### NCERT Exemplar Class 11 Mathematics

NCERT Class 11th Mathematics lays a strong foundation for the aspirants preparing for competitive examinations like JEE, NEET, KVPY, WBJEE and many other engineering and medical-related examinations. To score well and be in the top percentile in these examinations, students must have a clear-cut understanding of every concept covered in Class 11 Mathematics.

To excel in it, students need to do rigorous practice. Hence, they must have access to NCERT-related questions. NCERT Exemplar Class 11th Mathematics thus has all NCERT-based questions with solutions. . Students can find each topic covered with various types of questions with varying difficulty levels that they will face in the examination.

Moreover, they will also find tricks to solve these questions in less time with greater accuracy and will become more proficient in their calculations. The subject matter experts have designed the book after analysing the past year’s papers and the competitive examinations. The challenging questions will aid in developing the strong mindsets among the students, and as a result, they will start thinking more rationally. Thus, the book builds students’ approach, making them capable of solving difficult questions with ease and becoming more confident in the process.

Key Features of NCERT Solutions for Class 11 Mathematics Chapter 9

The more you revise, the more you retain. Hence, Extramarks NCERT Solutions for Class 11 Mathematics Chapter 9 provides a complete revision guide to every student irrespective of their level. Along with the revision guides, students will also get access to study notes, solved questions from NCERT textbook and Exemplars, and so on.

The key features are as follows:

• The entire chapter has been summarised in a point-wise manner in our NCERT solutions.
• Our NCERT solutions are prepared by subject matter experts working conscientiously and diligently to prepare authentic, concise answers which students can trust and enjoy the process of learning.
• All the important formulas are listed in a structured way for the students to revise quickly.
• After completing the NCERT Solutions for Class 11 Mathematics Chapter 9, students will be able to apply all the concepts related to sequence and series in other chapters.

Q.1 Write the first five terms of the sequence whose nth terms is:
an = n(n+2)

Ans

$\begin{array}{l}\text{We have}\\ {\text{a}}_{\text{n}}=\text{n}\left(\text{n}+\text{2}\right)\\ \mathrm{Substituting}\text{n}=1,2,3,4,5\text{respectively, we get}\\ {\text{a}}_{\text{1}}=\text{1}\left(\text{1}+\text{2}\right)\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=3\\ {\text{a}}_{\text{2}}=\text{2}\left(\text{2}+\text{2}\right)\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=8\\ {\text{a}}_{\text{3}}=\text{3}\left(\text{3}+\text{2}\right)\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=15\\ {\text{a}}_{\text{4}}=\text{4}\left(\text{4}+\text{2}\right)\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=24\\ {\text{a}}_{\text{5}}=\text{5}\left(\text{5}+\text{2}\right)\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=35\\ \mathrm{Therefore},\text{the first five terms are: 3, 8, 15, 24, 35.}\end{array}$

Q.2

$\begin{array}{l}\mathbf{\text{Write the first five terms of the sequence whose nth terms is:}}\\ {\mathbf{\text{a}}}_{\mathbf{\text{n}}}\mathbf{\text{=}}\frac{\mathbf{\text{n}}}{\mathbf{\text{n+1}}}\mathbf{\text{.}}\end{array}$

Ans

$\begin{array}{l}\text{We have}\\ {\mathrm{a}}_{\mathrm{n}}=\frac{\mathrm{n}}{\mathrm{n}+1}\\ \mathrm{Substituting}\text{n}=1,2,3,4,5\text{respectively, we get}\\ {\text{a}}_{\text{1}}=\frac{1}{1+1}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\frac{1}{2}\\ {\text{a}}_{\text{2}}=\frac{2}{2+1}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\frac{2}{3}\\ {\text{a}}_{\text{3}}=\text{}\frac{3}{3+1}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\frac{3}{4}\\ {\text{a}}_{\text{4}}=\frac{4}{4+1}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\frac{4}{5}\\ {\text{a}}_{\text{5}}=\frac{5}{5+1}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\frac{5}{6}\\ \mathrm{Therefore},\text{the first five terms are:}\frac{1}{2},\text{\hspace{0.17em}}\frac{2}{3},\frac{3}{4},\frac{4}{5},\frac{5}{6}.\end{array}$

Q.3 Write the first five terms of the sequence whose nth terms is:
an = 2n

Ans

${\text{We have, a}}_{\text{n}}={\text{2}}^{\text{n}}$ $\begin{array}{l}\mathrm{Substituting}\text{n}=1,2,3,4,5\text{respectively, we get}\\ {\text{a}}_{\text{1}}={\text{2}}^{\text{1}}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=2\\ {\text{a}}_{\text{2}}={\text{2}}^{\text{2}}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=4\\ {\text{a}}_{\text{3}}={\text{2}}^{\text{3}}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=8\\ {\text{a}}_{\text{4}}={\text{2}}^{\text{4}}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=16\\ {\text{a}}_{\text{5}}={\text{2}}^{\text{5}}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=32\\ \mathrm{Therefore},\text{the first five terms are: 2, 4, 8, 16, 32.}\end{array}$

Q.4

$\begin{array}{l}\mathbf{\text{Write the first five terms of the sequence whose nth terms is:}}\\ {\mathbf{\text{a}}}_{\mathbf{\text{n}}}\mathbf{\text{=}}\frac{\mathbf{\text{2n}}\mathbf{-}\mathbf{\text{3}}}{\mathbf{\text{6}}}\mathbf{\text{.}}\end{array}$

Ans

$\begin{array}{l}\text{We have}\\ {\mathrm{a}}_{\mathrm{n}}=\frac{2\mathrm{n}-3}{6}\\ \mathrm{Substituting}\text{n}=1,2,3,4,5\text{respectively, we get}\\ {\text{a}}_{\text{1}}=\frac{2\left(1\right)-3}{6}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=-\frac{1}{6}\\ {\text{a}}_{\text{2}}=\frac{2\left(2\right)-3}{6}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\frac{1}{6}\end{array}$ $\begin{array}{l}{\text{a}}_{\text{3}}=\frac{2\left(3\right)-3}{6}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\frac{3}{6}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\frac{1}{2}\\ {\text{a}}_{\text{4}}=\frac{2\left(4\right)-3}{6}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\frac{5}{6}\\ {\text{a}}_{\text{5}}=\frac{2\left(5\right)-3}{6}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\frac{7}{6}\\ \mathrm{Therefore},\text{the first five terms are:}-\frac{1}{6},\frac{1}{6},\frac{1}{2},\frac{5}{6},\frac{7}{6}.\end{array}$

Q.5 Write the first five terms of the sequence whose nth terms is an = (– 1)n–1 5n+1.

Ans

$\begin{array}{l}{\text{We have a}}_{\text{n}}={\left(-\text{1}\right)}^{\text{n}-\text{1}}{\text{5}}^{\text{n}+\text{1}}\\ \mathrm{Substituting}\text{n}=1,2,3,4,5\text{respectively, we get}\\ {\text{a}}_{\text{1}}={\left(–\text{1}\right)}^{\text{1}-\text{1}}{\text{5}}^{\text{1}+\text{1}}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=25\\ {\text{a}}_{\text{2}}={\left(–\text{1}\right)}^{\text{2}-\text{1}}{\text{5}}^{\text{2}+\text{1}}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=-125\\ {\text{a}}_{\text{3}}={\left(-\text{1}\right)}^{\text{3}-\text{1}}{\text{5}}^{\text{3}+\text{1}}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=625\end{array}$ $\begin{array}{l}{\text{a}}_{\text{4}}={\left(-\text{1}\right)}^{\text{4}-\text{1}}{\text{5}}^{\text{4}+\text{1}}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=-3125\\ {\text{a}}_{\text{5}}={\left(-\text{1}\right)}^{\text{5}-\text{1}}{\text{5}}^{\text{5}+\text{1}}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=15625\\ \mathrm{Therefore},\text{the first five terms are:25,}-\text{125,625,}-3125,15625.\end{array}$

Q.6

$\begin{array}{l}\mathbf{\text{Write the first five terms of the sequence whose nth terms is}}\\ {\mathbf{\text{a}}}_{\mathbf{\text{n}}}\mathbf{\text{=n}}\frac{{\mathbf{\text{n}}}^{\mathbf{\text{2}}}\mathbf{\text{+5}}}{\mathbf{\text{4}}}\end{array}$

Ans

$\begin{array}{l}\text{We have}{\mathrm{a}}_{\mathrm{n}}=\mathrm{n}\frac{{\mathrm{n}}^{2}+5}{4}\\ \text{Substituting n = 1, 2, 3, 4, 5 respectively, we get}\\ {\text{a}}_{\text{1}}=\left(1\right)\frac{{\left(1\right)}^{2}+5}{4}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\frac{6}{4}=\frac{3}{2}\\ {\text{a}}_{\text{2}}=\left(2\right)\frac{{\left(2\right)}^{2}+5}{4}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\frac{9}{2}\\ {\text{a}}_{\text{3}}=\left(3\right)\frac{{\left(3\right)}^{2}+5}{4}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\frac{42}{4}=\frac{21}{2}\\ {\text{a}}_{\text{4}}=\left(4\right)\frac{{\left(4\right)}^{2}+5}{4}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=21\\ {\text{a}}_{\text{5}}=\left(5\right)\frac{{\left(5\right)}^{2}+5}{4}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\frac{150}{4}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\frac{75}{2}\\ \text{Therefore, the first five terms are:}\frac{3}{2},\text{\hspace{0.17em}\hspace{0.17em}}\frac{9}{2},\text{\hspace{0.17em}\hspace{0.17em}}\frac{21}{2},\text{\hspace{0.17em}\hspace{0.17em}}21,\text{\hspace{0.17em}\hspace{0.17em}}\frac{75}{2}.\end{array}$

Q.7

$\begin{array}{l}\mathbf{\text{Find the indicated terms in each of the sequences given below}}\\ {\mathbf{\text{whose n}}}^{\mathbf{\text{th}}}\mathbf{\text{terms are:}}\\ \mathbf{1}\mathbf{.}\mathbf{\text{}}{\mathbf{a}}_{\mathbf{n}}\mathbf{=}\mathbf{\text{}}\mathbf{4}\mathbf{n}\mathbf{–}\mathbf{3}\mathbf{;}\mathbf{\text{}}{\mathbf{a}}_{\mathbf{17}}\mathbf{,}\mathbf{\text{}}{\mathbf{a}}_{\mathbf{24}}\\ {\mathbf{\text{2. a}}}_{\mathbf{\text{n}}}\mathbf{\text{=}}\frac{{\mathbf{\text{n}}}^{\mathbf{\text{2}}}}{{\mathbf{\text{2}}}^{\mathbf{\text{n}}}}{\mathbf{\text{;\hspace{0.17em}\hspace{0.17em}a}}}_{\mathbf{\text{7}}}\\ \mathbf{3}\mathbf{.}\mathbf{\text{}}{\mathbf{a}}_{\mathbf{n}}\mathbf{=}\mathbf{\text{}}{\left(–1\right)}^{\mathbf{n}\mathbf{–}\mathbf{1}}{\mathbf{n}}^{\mathbf{3}}\mathbf{;}\mathbf{\text{}}{\mathbf{a}}_{\mathbf{9}}\\ {\mathbf{\text{4. a}}}_{\mathbf{\text{n}}}\mathbf{\text{=}}\frac{\mathbf{\text{n}}\left(\text{n}-\text{2}\right)}{\mathbf{\text{n+3}}}{\mathbf{\text{; a}}}_{\mathbf{\text{20}}}\end{array}$

Ans

1.

$\begin{array}{l}\text{Given:}\\ {\text{a}}_{\text{n}}=\text{4n}–\text{3}\\ \mathrm{Substituting}\text{n=17 and 24 respectively, we get}\\ {\text{a}}_{\text{17}}=\text{4}\left(17\right)-\text{3}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\text{68}-\text{3}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\text{65}\\ {\text{a}}_{\text{24}}=\text{4}\left(24\right)-\text{3}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=96-3\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=93\end{array}$

2.

$\begin{array}{l}\text{Given:\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}{\mathrm{a}}_{\mathrm{n}}=\frac{{\mathrm{n}}^{2}}{{2}^{\mathrm{n}}}\\ \mathrm{Substituting}\text{}\mathrm{n}=7\text{, we get}\\ {\text{a}}_{\text{7}}=\frac{{7}^{2}}{{2}^{7}}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\frac{49}{128}\\ {\text{Thus, the value of a}}_{\text{7}}\text{is}\frac{49}{128}\text{.}\end{array}$

3.

$\begin{array}{l}{\text{Given:\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}a}}_{\text{n}}={\left(-\text{1}\right)}^{\text{n}–\text{1}}{\text{n}}^{\text{3}}\\ \mathrm{Substituting}\text{}\mathrm{n}=9\text{, we get}\\ {\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}a}}_{\text{9}}={\left(-\text{1}\right)}^{\text{9}–\text{1}}{\left(9\right)}^{\text{3}}\\ ={\left(-\text{1}\right)}^{8}\left(729\right)\\ =729\\ \mathrm{Thus},{\text{the value of a}}_{\text{9}}\text{is 729.}\end{array}$ $\begin{array}{l}\text{4.}\end{array}$ $\begin{array}{l}\text{Given:\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}{\mathrm{a}}_{\mathrm{n}}=\frac{\mathrm{n}\left(\mathrm{n}-2\right)}{\mathrm{n}+3}\\ \mathrm{Substituting}\text{}\mathrm{n}=20\text{, we get}\\ {\text{\hspace{0.17em}\hspace{0.17em}a}}_{\text{20}}=\frac{20\left(20-2\right)}{20+3}\end{array}$ $\begin{array}{l}\text{}\text{}\text{}\end{array}$ $\begin{array}{l}=\frac{20\left(18\right)}{23}\\ =\frac{360}{23}\\ \mathrm{Thus},{\text{the value of a}}_{\text{20}}\text{is}\frac{360}{23}\text{.}\end{array}$

Q.8

$\begin{array}{l}\mathbf{\text{Write the first five terms of each of the sequences given below and}}\\ \mathbf{\text{obtain the corresponding series:}}\\ {\mathbf{\text{1. a}}}_{\mathbf{\text{1}}}\mathbf{\text{=}}\mathbf{3}{\mathbf{\text{, a}}}_{\mathbf{\text{n}}}{\mathbf{\text{=3a}}}_{\mathbf{\text{n-1}}}\mathbf{\text{+2, for all n}}\mathbf{>}\mathbf{1}\\ {\mathbf{\text{2. a}}}_{\mathbf{\text{1}}}\mathbf{\text{=}}\mathbf{–}{\mathbf{\text{1, a}}}_{\mathbf{\text{n}}}\mathbf{\text{=}}\frac{{\mathbf{\text{a}}}_{\mathbf{\text{n-1}}}}{\mathbf{\text{n}}}\mathbf{\text{, n}}\mathbf{\ge }\mathbf{\text{2}}\\ \mathbf{3}\mathbf{.}\mathbf{\text{}}{\mathbf{a}}_{\mathbf{1}}\mathbf{=}{\mathbf{a}}_{\mathbf{2}}\mathbf{=}\mathbf{2}\mathbf{,}\mathbf{\text{}}{\mathbf{a}}_{\mathbf{n}}\mathbf{=}{{\mathbf{a}}_{\mathbf{n}}}_{\mathbf{–}\mathbf{1}}\mathbf{–}\mathbf{1}\mathbf{,}\mathbf{\text{}}\mathbf{n}\mathbf{\text{}}\mathbf{>}\mathbf{\text{}}\mathbf{2}\end{array}$

Ans

1.

$\begin{array}{l}\mathrm{Given}:{\mathrm{a}}_{\mathrm{n}}=3{\mathrm{a}}_{\mathrm{n}}-1+2{\text{and a}}_{\text{1}}=\text{3}\\ \mathrm{Substituting}\text{n}=\text{2, we get}\\ {\mathrm{a}}_{2}=3{{\mathrm{a}}_{2}}_{-1}+2\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=3{\mathrm{a}}_{1}+2\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=3\left(3\right)+2\left[\mathrm{Putting}{\text{a}}_{\text{1}}=\text{3}\right]\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=11\\ \mathrm{Substituting}\text{n}=\text{3, we get}\\ {\mathrm{a}}_{3}=3{{\mathrm{a}}_{3}}_{-1}+2\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=3{\mathrm{a}}_{2}+2\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=3\left(11\right)+2\left[\mathrm{Putting}{\text{a}}_{\text{2}}=\text{11}\right]\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=35\\ \mathrm{Substituting}\text{n}=\text{4, we get}\\ {\mathrm{a}}_{4}=3{{\mathrm{a}}_{4}}_{-1}+2\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=3{\mathrm{a}}_{3}+2\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=3\left(35\right)+2\left[\mathrm{Putting}{\text{a}}_{\text{3}}=\text{35}\right]\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=107\end{array}$ $\begin{array}{l}\mathrm{Substituting}\text{n}=\text{5, we get}\\ {\mathrm{a}}_{5}=3{{\mathrm{a}}_{5}}_{-1}+2\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=3{\mathrm{a}}_{4}+2\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=3\left(107\right)+2\left[\mathrm{Putting}{\text{a}}_{\text{4}}=\text{107}\right]\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=323\\ \mathrm{Thus},\text{the first five terms are 3, 11, 35, 107 and 323.}\\ \text{Corresponding series is:\hspace{0.17em} 3}+\text{11}+\text{35}+\text{107}+\text{323}+...\end{array}$

2.

$\begin{array}{l}\mathrm{Given}:{\mathrm{a}}_{1}=-1,{\mathrm{a}}_{\mathrm{n}}=\frac{{\mathrm{a}}_{\mathrm{n}–1}}{\mathrm{n}},\mathrm{n}\ge 2\\ \mathrm{Substituting}\text{n}=\text{2, we get}\\ {\mathrm{a}}_{2}=\frac{{\mathrm{a}}_{2–1}}{2}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\frac{{\mathrm{a}}_{1}}{2}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\frac{-1}{2}\left[\mathrm{Putting}{\text{a}}_{\text{1}}=-\text{1}\right]\\ \mathrm{Substituting}\text{n}=\text{3, we get}\\ {\mathrm{a}}_{3}=\frac{{\mathrm{a}}_{3–1}}{3}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\frac{{\mathrm{a}}_{2}}{3}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\frac{\left(\frac{-1}{2}\right)}{3}\left[\mathrm{Putting}{\text{a}}_{\text{2}}=\frac{-1}{2}\right]\end{array}$ $\begin{array}{l}\text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\end{array}$ $\begin{array}{l}=\frac{-1}{6}\\ \mathrm{Substituting}\text{n}=\text{4, we get}\\ {\mathrm{a}}_{4}=\frac{{\mathrm{a}}_{4–1}}{4}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\frac{{\mathrm{a}}_{3}}{4}\end{array}$ $\begin{array}{l}\text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\end{array}$ $\begin{array}{l}=\frac{\left(\frac{-1}{6}\right)}{4}\left[\mathrm{Putting}{\text{a}}_{\text{3}}=\frac{-1}{6}\right]\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\frac{-1}{24}\\ \mathrm{Substituting}\text{n}=\text{5, we get}\\ {\mathrm{a}}_{5}=\frac{{\mathrm{a}}_{5–1}}{5}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\frac{{\mathrm{a}}_{4}}{5}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\frac{\left(\frac{-1}{24}\right)}{5}\left[\mathrm{Putting}{\text{a}}_{\text{4}}=\frac{-1}{24}\right]\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\frac{-1}{120}\\ \mathrm{Thus},\text{the first five terms are}-\text{1,}\frac{-\text{1}}{2}\text{,}\frac{-\text{1}}{6}\text{,}\frac{-\text{1}}{24}\text{and}\frac{-\text{1}}{120}\text{.}\\ \text{The corresponding series is:}\\ \left(-\text{1}\right)+\left(\frac{-\text{1}}{2}\right)+\left(\frac{-\text{1}}{6}\right)+\left(\frac{-\text{1}}{24}\right)+\left(\frac{-\text{1}}{120}\right)+...\end{array}$

3.

$\begin{array}{l}\mathrm{Given}:{\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}a}}_{\text{n}}={{\text{a}}_{\text{n}}}_{–\text{1}}-\text{1},\text{\hspace{0.17em}\hspace{0.17em}}{\mathrm{a}}_{1}={\mathrm{a}}_{2}=2,\text{n}>\text{2}\\ \mathrm{Substituting}\text{n}=\text{3, we get}\\ {\mathrm{a}}_{3}={{\text{a}}_{\text{3}}}_{–\text{1}}-\text{1}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}={\text{a}}_{\text{2}}-\text{1}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=2-1\left[\mathrm{Putting}{\text{a}}_{\text{2}}=2\right]\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=1\end{array}$ $\begin{array}{l}\end{array}$ $\begin{array}{l}\mathrm{Substituting}\text{n}=\text{4, we get}\\ {\mathrm{a}}_{4}={{\text{a}}_{\text{4}}}_{–\text{1}}-\text{1}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}={\text{a}}_{\text{3}}-\text{1}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=1-1\left[\mathrm{Putting}{\text{a}}_{\text{3}}=1\right]\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=0\\ \mathrm{Substituting}\text{n}=\text{5, we get}\\ {\mathrm{a}}_{5}={{\text{a}}_{\text{4}}}_{–\text{1}}-\text{1}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}={\text{a}}_{\text{3}}-\text{1}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=0-1\left[\mathrm{Putting}{\text{a}}_{\text{4}}=0\right]\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=-1\\ \mathrm{Thus},\text{the first five terms of series are:}\\ \text{2,\hspace{0.17em}2,\hspace{0.17em}1,\hspace{0.17em}0,}-\text{1}\\ \text{The series is:\hspace{0.17em}\hspace{0.17em}2}+\text{\hspace{0.17em}2}+\text{\hspace{0.17em}1}+\text{\hspace{0.17em}0}+\left(-\text{1}\right)+...\end{array}$

Q.9

$\begin{array}{l}{\mathbf{\text{The Fibonacci sequence is defined by 1=a}}}_{\mathbf{\text{1}}}{\mathbf{\text{=a}}}_{\mathbf{\text{2}}}{\mathbf{\text{and a}}}_{\mathbf{\text{n}}}{\mathbf{\text{=a}}}_{\mathbf{\text{n}}\mathbf{-}\mathbf{\text{1}}}{\mathbf{\text{+a}}}_{\mathbf{\text{n}}\mathbf{-}\mathbf{\text{2}}}\mathbf{\text{,\hspace{0.17em}\hspace{0.17em}n>2.}}\\ \mathbf{\text{Find}}\frac{{\mathbf{\text{a}}}_{\mathbf{\text{n+1}}}}{{\mathbf{\text{a}}}_{\mathbf{\text{n}}}}\mathbf{\text{, for n=1,2,3,4,5.}}\end{array}$

Ans

$\begin{array}{l}\text{TheFibonaccisequenceisdefinedby}\\ {\text{1=a}}_{\text{1}}{\text{=a}}_{\text{2}}{\text{anda}}_{\text{n}}{\text{=a}}_{\text{n}-\text{1}}{\text{+a}}_{\text{n}-\text{2}}\text{,n>2.}\\ \text{Substitutingn=3,weget}\\ \text{}{\mathrm{a}}_{3}={\mathrm{a}}_{2}+{\mathrm{a}}_{1}\\ \text{}=1+1\\ \text{}=2\\ \text{Forn=1,weget}\\ \text{}\frac{{\text{a}}_{\text{1+1}}}{{\text{a}}_{\text{1}}}\text{=}\frac{{\text{a}}_{\text{2}}}{{\text{a}}_{\text{1}}}\\ \text{=}\frac{\text{1}}{\text{1}}\\ \text{}\frac{{\text{a}}_{\text{2}}}{{\text{a}}_{\text{1}}}\text{=1}\\ \text{and}\frac{{\mathrm{a}}_{3}}{{\mathrm{a}}_{2}}=\frac{2}{1}=2\\ \text{}\frac{{\text{a}}_{\text{n+1}}}{{\text{a}}_{\text{n}}}\text{=}\frac{{\text{a}}_{\text{n}}{\text{+a}}_{\text{n}-\text{1}}}{{\text{a}}_{\text{n-1}}{\text{+a}}_{\text{n}-\text{2}}}\\ \text{Substituting}\mathrm{n}=3\text{,weget}\\ \frac{{\text{a}}_{\text{3+1}}}{{\text{a}}_{\text{3}}}\text{=}\frac{{\text{a}}_{\text{3}}{\text{+a}}_{\text{3}-\text{1}}}{{\text{a}}_{\text{3}-\text{1}}{\text{+a}}_{\text{3}-\text{2}}}\\ \text{}\frac{{\text{a}}_{\text{4}}}{{\text{a}}_{\text{3}}}\text{=}\frac{{\text{a}}_{\text{3}}{\text{+a}}_{\text{2}}}{{\text{a}}_{\text{2}}{\text{+a}}_{\text{1}}}\\ \text{=}\frac{\left({\text{a}}_{\text{2}}{\text{+a}}_{\text{1}}\right){\text{+a}}_{\text{2}}}{{\text{a}}_{\text{2}}{\text{+a}}_{\text{1}}}\text{}\left[{\mathrm{a}}_{3}={\mathrm{a}}_{2}+{\mathrm{a}}_{1}\right]\\ \text{=}\frac{\left(\text{1+1}\right)\text{+1}}{\text{1+1}}\\ \text{=}\frac{\text{3}}{\text{2}}\\ \mathrm{Substituting}\text{n}=4,\text{we get}\\ {\text{a}}_{\text{4}}={\text{a}}_{3}+{\text{a}}_{2}\\ \text{}=\text{}2+1\\ \text{}=3\\ \frac{{\text{a}}_{4+1}}{{\text{a}}_{4}}=\frac{{\text{a}}_{4}+{\text{a}}_{3}}{{\text{a}}_{3}+{\text{a}}_{2}}\\ \text{}\frac{{\text{a}}_{5}}{{\text{a}}_{4}}=\frac{3+2}{2+1}\\ \text{}=\frac{5}{3}\\ \mathrm{Substituting}\text{n}=5,\text{weget}\\ {\text{a}}_{\text{5}}={\text{a}}_{4}+{\text{a}}_{3}\\ \text{}=3+2\\ \text{}=5\\ \text{}\frac{{\text{a}}_{5+1}}{{\text{a}}_{5}}=\frac{{\text{a}}_{5}+{\text{a}}_{4}}{{\text{a}}_{4}+{\text{a}}_{3}}\\ \text{}\frac{{\text{a}}_{6}}{{\text{a}}_{5}}=\frac{5+3}{3+2}\\ \text{}=\frac{8}{5}\\ \mathrm{Thus},\text{therequiredtermsare:}1,2,\frac{3}{2},\frac{5}{3},\frac{8}{5}.\end{array}$

Q.10 Find the sum of odd integers from 1 to 2001.

Ans

$\begin{array}{l}\mathrm{The}\text{odd integers from 1 to 2001 are:}\\ 1,\text{\hspace{0.17em}}3,5,7,9,11,13,...,1999,2001\\ \mathrm{a}=1,\text{\hspace{0.17em}}\mathrm{d}=3-1=2\\ \text{\hspace{0.17em}\hspace{0.17em}}\mathrm{l}=\mathrm{a}+\left(\mathrm{n}-1\right)\mathrm{d}\\ \text{\hspace{0.17em}}2001=1+\left(\mathrm{n}-1\right)2\\ 2001-1=\left(\mathrm{n}-1\right)2\\ \mathrm{n}-1=\frac{2000}{2}\\ \text{}\mathrm{n}=1000+1\\ =1001\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}{\mathrm{S}}_{\mathrm{n}}=\frac{\mathrm{n}}{2}\left(\mathrm{a}+\mathrm{l}\right)\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}{\mathrm{S}}_{1001}=\frac{1001}{2}\left(1+2001\right)\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}{\mathrm{S}}_{1001}=\frac{1001}{2}\left(2002\right)\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=1001\left(1001\right)\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=1002001\\ \mathrm{Thus},\text{the sum of odd integers from 1 to 2001 is 1002001.}\end{array}$

Q.11 Find the sum of all natural numbers lying between 100 and 1000, which are multiples of 5.

Ans

$\begin{array}{l}\mathrm{The}\text{multiples of 5 from 100 to 1000 are:}\\ 105,\text{\hspace{0.17em}}105,110,\text{\hspace{0.17em}}115,\text{\hspace{0.17em}}120,\text{\hspace{0.17em}\hspace{0.17em}}125,...,995\\ \mathrm{a}=100,\text{\hspace{0.17em}}\mathrm{d}=105-100=5\\ \mathrm{l}=\mathrm{a}+\left(\mathrm{n}-1\right)\mathrm{d}\\ 995=105+\left(\mathrm{n}-1\right)5\\ 995-105=\left(\mathrm{n}-1\right)5\\ \mathrm{n}-1=\frac{890}{5}\\ \text{}\mathrm{n}=178+1\\ \text{}=179\\ \text{}{\mathrm{S}}_{\mathrm{n}}=\frac{\mathrm{n}}{2}\left(\mathrm{a}+\mathrm{l}\right)\\ \text{}{\mathrm{S}}_{179}=\frac{179}{2}\left(105+995\right)\\ =\frac{179}{2}×\left(1100\right)\\ =98450\\ \mathrm{Thus},\text{the sum of the natural numbers which are multiple of 5}\\ \text{between 100 to 1000 is}98450.\end{array}$

Q.12 In an A.P., the first term is 2 and the sum of the first five terms is one-fourth of the next five terms. Show that 20th term is –112.

Ans

$\begin{array}{l}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{First}\text{term of A.P.}=\text{2}\\ \text{Let common difference}=\mathrm{d}\\ \mathrm{According}\text{to given condition:}\\ \text{Sum of first 5 terms}=\frac{1}{4}\left(\mathrm{Sum}\text{of next 5 terms}\right)\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}{\mathrm{S}}_{5}=\frac{1}{4}\left({\mathrm{S}}_{10}-{\mathrm{S}}_{5}\right)\\ ⇒\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}4{\mathrm{S}}_{5}={\mathrm{S}}_{10}-{\mathrm{S}}_{5}\\ ⇒\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}5{\mathrm{S}}_{5}={\mathrm{S}}_{10}\end{array}$ $\begin{array}{l}⇒5\left[\frac{5}{2}\left\{2×2+\left(5-1\right)\mathrm{d}\right\}\right]=\left[\frac{10}{2}\left\{2×2+\left(10-1\right)\mathrm{d}\right\}\right]\\ \left[âˆµ{\mathrm{S}}_{\mathrm{n}}=\frac{\mathrm{n}}{2}\left\{2\mathrm{a}+\left(\mathrm{n}-1\right)\mathrm{d}\right\}\right]\\ ⇒\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\frac{25}{2}\left(4+4\mathrm{d}\right)=5\left(4+9\mathrm{d}\right)\\ ⇒\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}25\left(2+2\mathrm{d}\right)=5\left(4+9\mathrm{d}\right)\\ ⇒\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}50+50\mathrm{d}=20+45\mathrm{d}\\ ⇒\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}5\mathrm{d}=20-50\\ ⇒\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{d}=-\frac{30}{5}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=-6\\ \mathrm{Now},\text{sum of 20 terms.}\\ {\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}T}}_{\text{20}}=2+\left(20-1\right)×-6\left[âˆµ{\mathrm{a}}_{\mathrm{n}}=\mathrm{a}+\left(\mathrm{n}-1\right)\mathrm{d}\right]\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=2-114\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=-112\\ \mathrm{Thus},{\text{the 20}}^{\text{th}}\text{term is}-\text{112.}\end{array}$

Q.13

$\begin{array}{l}\mathbf{\text{How many terms of the A.P.\hspace{0.17em}\hspace{0.17em}}}\mathbf{-}\mathbf{\text{6,}}\mathbf{-}\frac{\mathbf{\text{11}}}{\mathbf{\text{2}}}\mathbf{\text{,}}\mathbf{-}\mathbf{\text{5,}}\mathbf{.}\mathbf{..}\mathbf{\text{\hspace{0.17em} are needed to give}}\\ \mathbf{\text{the sum}}\mathbf{-}\mathbf{\text{25?}}\end{array}$

Ans

$\begin{array}{l}\mathrm{The}\text{given A.P. is}-6,-\frac{11}{2},-5,...\text{\hspace{0.17em}}\\ \mathrm{Let}\text{the sum of n terms of A.P. is}-\text{25.Then,}\end{array}$ $\begin{array}{l}{\mathrm{S}}_{\mathrm{n}}=\frac{\mathrm{n}}{2}\left\{2\mathrm{a}+\left(\mathrm{n}-1\right)\mathrm{d}\right\}\\ \mathrm{a}=-6\\ \mathrm{d}=-\frac{11}{2}-\left(-6\right)\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=-\frac{11}{2}+6=\frac{1}{2}\\ -25=\frac{\mathrm{n}}{2}\left\{2×-6+\left(\mathrm{n}-1\right)×\frac{1}{2}\right\}\left[âˆµ{\mathrm{S}}_{\mathrm{n}}=\frac{\mathrm{n}}{2}\left\{2\mathrm{a}+\left(\mathrm{n}-1\right)\mathrm{d}\right\}\right]\\ -25=\frac{\mathrm{n}}{2}\left\{-12+\left(\mathrm{n}-1\right)×\frac{1}{2}\right\}\\ -100=\mathrm{n}\left(-24+\mathrm{n}-1\right)\\ -100=-25\text{\hspace{0.17em}}\mathrm{n}+{\mathrm{n}}^{2}\\ {\mathrm{n}}^{2}-25\text{\hspace{0.17em}}\mathrm{n}+100=0\\ \left(\mathrm{n}-20\right)\left(\mathrm{n}-5\right)=0\\ ⇒\text{\hspace{0.17em} \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{n}=20,\text{\hspace{0.17em}\hspace{0.17em}}5\\ \mathrm{Thus},\text{the number of terms of A.P. is 5 or 20.}\end{array}$

Q.14

$\begin{array}{l}{\mathbf{\text{In an A.P., if p}}}^{\mathbf{\text{th}}}\mathbf{\text{term is}}\frac{\mathbf{\text{1}}}{\mathbf{\text{p}}}{\mathbf{\text{and q}}}^{\mathbf{\text{th}}}\mathbf{\text{term is}}\frac{\mathbf{\text{1}}}{\mathbf{\text{q}}}\mathbf{\text{, prove that the sum of first}}\\ \mathbf{\text{pq terms is}}\frac{\mathbf{\text{1}}}{\mathbf{\text{2}}}\left(\text{pq+1}\right)\mathbf{\text{, where p}}\mathbf{\ne }\mathbf{\text{q.}}\end{array}$

Ans

$\begin{array}{l}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{Let}\text{first term of A.P.}=\mathrm{a}\\ \mathrm{Let}\text{common difference of A.P.}=\mathrm{d}\\ \mathrm{Since},{\text{T}}_{\text{p}}=\frac{1}{\mathrm{q}}\end{array}$ $\begin{array}{l}⇒\mathrm{a}+\left(\mathrm{p}-1\right)\mathrm{d}=\frac{1}{\mathrm{q}}...\left(\mathrm{i}\right)\\ \mathrm{And},{\text{T}}_{\text{q}}=\frac{1}{\mathrm{p}}\\ ⇒\mathrm{a}+\left(\mathrm{q}-1\right)\mathrm{d}=\frac{1}{\mathrm{p}}...\left(\mathrm{ii}\right)\\ \mathrm{Subtracting}\text{equation}\left(\mathrm{ii}\right)\text{from equation}\left(\mathrm{i}\right),\text{we get}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\left(\mathrm{p}-1-\mathrm{q}+1\right)\mathrm{d}=\frac{1}{\mathrm{q}}-\frac{1}{\mathrm{p}}\\ ⇒\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\left(\mathrm{p}-\mathrm{q}\right)\mathrm{d}=\frac{\mathrm{p}-\mathrm{q}}{\mathrm{pq}}\\ ⇒\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{d}=\frac{1}{\mathrm{pq}}\\ \mathrm{Substituting}\text{value of d in equation}\left(\mathrm{i}\right),\text{\hspace{0.17em}}\mathrm{we}\text{\hspace{0.17em}}\mathrm{get}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{a}+\left(\mathrm{p}-1\right)\frac{1}{\mathrm{pq}}=\frac{1}{\mathrm{q}}\\ ⇒\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{a}+\frac{\mathrm{p}}{\mathrm{pq}}-\frac{1}{\mathrm{pq}}=\frac{1}{\mathrm{q}}\\ ⇒\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{a}=\frac{1}{\mathrm{pq}}\\ \mathrm{Now},\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}{\mathrm{S}}_{\mathrm{pq}}=\frac{\mathrm{pq}}{2}\left\{2×\frac{1}{\mathrm{pq}}+\left(\mathrm{pq}-1\right)\frac{1}{\mathrm{pq}}\right\}\\ \left[âˆµ{\mathrm{S}}_{\mathrm{n}}=\frac{\mathrm{n}}{2}\left\{2\mathrm{a}+\left(\mathrm{n}-1\right)\mathrm{d}\right\}\right]\end{array}$ $\begin{array}{l}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{}\text{}\text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\end{array}$ $\begin{array}{l}\text{\hspace{0.17em}}{\mathrm{S}}_{\mathrm{pq}}=\frac{\mathrm{pq}}{2}\left\{2×\frac{1}{\mathrm{pq}}+\left(\mathrm{pq}-1\right)\frac{1}{\mathrm{pq}}\right\}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\frac{\mathrm{pq}}{2}\left(\frac{2}{\mathrm{pq}}+1-\frac{1}{\mathrm{pq}}\right)\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\frac{\mathrm{pq}}{2}\left(\frac{1}{\mathrm{pq}}+1\right)\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\frac{1}{2}\left(1+\mathrm{pq}\right)\\ \mathrm{Thus},\text{the sum of pq terms is}\frac{1}{2}\left(\mathrm{pq}+1\right).\end{array}$

Q.15 If the sum of a certain number of terms of the A.P. 25, 22, 19, … is 116. Find the last term.

Ans

$\begin{array}{l}\begin{array}{l}\mathrm{Given}\text{A.P. 25},\text{22},\text{19},\dots \\ \mathrm{First}\text{term of A.P.}=\text{25}\\ \text{Common difference of A.P.}=22-25\end{array}\\ \begin{array}{l}\text{}=-3\\ \mathrm{Let}\text{sum of n terms of A.P.}=116\end{array}\\ \text{i}.\text{e}.,{\text{S}}_{\text{n}}=116\\ \frac{\text{n}}{2}\left\{2×25+\left(\text{n}-1\right)\left(-3\right)\right\}=116\\ ⇒\text{n}\left(50-3\text{n}+3\right)=232\\ \begin{array}{l}⇒\text{}\left(53\text{n}-3{\text{n}}^{2}\right)=232\\ ⇒\text{}3{\text{n}}^{2}-53\text{n}+232=0\end{array}\\ ⇒\text{}3{\text{n}}^{2}-24\text{n}-29\text{n}+232=0\\ ⇒\text{}3\text{n}\left(\text{n}-8\right)-29\left(\text{n}-8\right)=0\\ \begin{array}{l}⇒\text{}\left(\text{n}-8\right)\left(3\text{n}-29\right)=0\\ ⇒\text{n}=8,\frac{29}{3}\end{array}\\ \mathrm{Since},\text{n is a natural number.So,}\\ \begin{array}{l}\text{n}=\text{8}\\ \text{Last term of A.P.}=\text{a}+\left(\text{n}-1\right)\text{d}\end{array}\\ \text{}=25+\left(8-1\right)\left(-3\right)\\ \text{}=25-21\\ \text{}=4\\ \mathrm{Thus},\text{the last term of A.P. is 4.}\end{array}$

Q.16 Find the sum to n terms of the A.P., whose kth term is 5k + 1.

Ans

$\begin{array}{l}{\mathrm{k}}^{\mathrm{th}}\text{term of A.P.}=\text{5k+1}\\ ⇒\text{\hspace{0.17em}}{\mathrm{T}}_{\mathrm{k}}=\text{5k+1}\\ \mathrm{Substituting}\text{k}=1,2,3,...\text{respectively.}\\ {\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}T}}_{\text{1}}=5\left(1\right)+1\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=6\\ {\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}T}}_{\text{2}}=5\left(2\right)+1\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=11\\ {\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}T}}_{\text{3}}=5\left(3\right)+1\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=16\\ \mathrm{Common}\text{difference}={\mathrm{T}}_{2}-{\mathrm{T}}_{1}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=11-6\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=5\end{array}$ $\begin{array}{l}\mathrm{Sum}{\text{of n terms,S}}_{\text{n}}=\frac{\mathrm{n}}{2}\left\{2\mathrm{a}+\left(\mathrm{n}-1\right)\mathrm{d}\right\}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\frac{\mathrm{n}}{2}\left\{2×6+\left(\mathrm{n}-1\right)5\right\}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\frac{\mathrm{n}}{2}\left(12+5\mathrm{n}-5\right)\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\frac{\mathrm{n}}{2}\left(7+5\mathrm{n}\right)\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\frac{\mathrm{n}}{2}\left(5\mathrm{n}+7\right)\\ \mathrm{Thus},\text{the sum of n terms is}\frac{\mathrm{n}}{2}\left(5\mathrm{n}+7\right).\end{array}$

Q.17 If the sum of n terms of an A.P. is (pn + qn2), where p and q are constants, find the common difference.

Ans

$\begin{array}{l}\mathrm{Given},\text{\hspace{0.17em}}{\mathrm{S}}_{\mathrm{n}}=\left(\mathrm{pn}+{\mathrm{qn}}^{2}\right)\\ \mathrm{Putting}\text{n}=1,2,3\text{\hspace{0.17em}}\mathrm{respectively},\mathrm{we}\text{\hspace{0.17em}}\mathrm{get}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}{\mathrm{S}}_{1}=\left(\mathrm{p}×1+\mathrm{q}×{1}^{2}\right)\\ \text{\hspace{0.17em}}=\mathrm{p}+\mathrm{q}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}{\mathrm{S}}_{2}=\left(\mathrm{p}×2+\mathrm{q}×{2}^{2}\right)\\ \text{\hspace{0.17em}}=2\text{\hspace{0.17em}}\mathrm{p}+4\text{\hspace{0.17em}}\mathrm{q}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}{\mathrm{S}}_{3}=\left(\mathrm{p}×3+\mathrm{q}×{3}^{2}\right)\\ \text{\hspace{0.17em}}=3\text{\hspace{0.17em}}\mathrm{p}+9\text{\hspace{0.17em}}\mathrm{q}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}{\mathrm{T}}_{1}={\mathrm{S}}_{2}-{\mathrm{S}}_{1}\\ \text{\hspace{0.17em}}=\left(2\text{\hspace{0.17em}}\mathrm{p}+4\text{\hspace{0.17em}}\mathrm{q}\right)-\left(\mathrm{p}+\mathrm{q}\right)\end{array}$ $\begin{array}{l}\text{\hspace{0.17em}}=\mathrm{p}+3\mathrm{q}\\ {\mathrm{T}}_{2}={\mathrm{S}}_{3}-{\mathrm{S}}_{2}\\ \text{\hspace{0.17em}}=\left(3\text{\hspace{0.17em}}\mathrm{p}+9\text{\hspace{0.17em}}\mathrm{q}\right)-\left(2\text{\hspace{0.17em}}\mathrm{p}+4\text{\hspace{0.17em}}\mathrm{q}\right)\\ \text{\hspace{0.17em}}=\mathrm{p}+5\text{\hspace{0.17em}}\mathrm{q}\\ \mathrm{Common}\text{difference,}\\ \text{d}={\mathrm{T}}_{2}-{\mathrm{T}}_{1}\\ \text{\hspace{0.17em}}=\left(\mathrm{p}+5\text{\hspace{0.17em}}\mathrm{q}\right)-\left(\mathrm{p}+3\mathrm{q}\right)\\ \text{\hspace{0.17em}}=2\mathrm{q}\\ \mathrm{Thus},\text{common difference is 2q.}\end{array}$

Q.18 The sums of n terms of two arithmetic progressions are in the ratio 5n + 4 : 9n + 6. Find the ratio of their 18th terms.

Ans

$\begin{array}{l}\mathrm{Since},\\ \frac{\mathrm{Sum}\text{of n terms of an A.P.}}{\mathrm{Sum}\text{of n terms of another A.P.}}=\frac{\text{5n}+\text{4}}{\text{9n}+\text{6}}\\ ⇒\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\frac{\frac{\mathrm{n}}{2}\left\{2\mathrm{a}+\left(\mathrm{n}-1\right)\mathrm{d}\right\}}{\frac{\mathrm{n}}{2}\left\{2\mathrm{A}+\left(\mathrm{n}-1\right)\mathrm{D}\right\}}=\frac{\text{5n}+\text{4}}{\text{9n}+\text{6}}\\ ⇒\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\frac{\left\{\mathrm{a}+\left(\frac{\mathrm{n}-1}{2}\right)\mathrm{d}\right\}}{\left\{\mathrm{A}+\left(\frac{\mathrm{n}-1}{2}\right)\mathrm{D}\right\}}=\frac{\text{5n}+\text{4}}{\text{9n}+\text{6}}\text{\hspace{0.17em} \hspace{0.17em}\hspace{0.17em}}...\left(\mathrm{i}\right)\\ \mathrm{Since},{\text{T}}_{\text{18}}=\text{a}+\text{17d. So, Substituting}\frac{\mathrm{n}-1}{2}=17\end{array}$ $\begin{array}{l}\mathrm{or}\text{\hspace{0.17em}}\mathrm{n}=35\text{in equation}\left(\mathrm{i}\right),\text{we get}\\ \frac{\left\{\mathrm{a}+17\mathrm{d}\right\}}{\left\{\mathrm{A}+17\mathrm{D}\right\}}=\frac{\text{5}×\text{35}+\text{4}}{\text{9}×\text{35}+\text{6}}\\ \frac{{\text{18}}^{\text{th}}\text{term of an A.P.}}{{\text{18}}^{\text{th}}\text{term of another A.P.}}=\frac{\text{175}+\text{4}}{\text{315}+\text{6}}\\ =\frac{179}{321}\\ \mathrm{Thus},{\text{the ratio of 18}}^{\text{th}}\text{terms of two A.P. is 179:321.}\end{array}$

Q.19 If the sum of first p terms of an A.P. is equal to the sum of the first q terms, then find the sum of the first (p + q) terms.

Ans

$\begin{array}{l}\text{first term of A.P.}=\text{a}\\ \mathrm{and}\text{common difference of A.P.}=\text{d}\\ \text{\hspace{0.17em}\hspace{0.17em}}{\mathrm{S}}_{\mathrm{p}}={\mathrm{S}}_{\mathrm{q}}\left(\mathrm{Given}\right)\\ ⇒\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\frac{\mathrm{p}}{2}\left\{2\mathrm{a}+\left(\mathrm{p}-1\right)\mathrm{d}\right\}=\frac{\mathrm{q}}{2}\left\{2\mathrm{a}+\left(\mathrm{q}-1\right)\mathrm{d}\right\}\\ ⇒\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}2\mathrm{ap}+\mathrm{p}\left(\mathrm{p}-1\right)\mathrm{d}=2\mathrm{aq}+\mathrm{q}\left(\mathrm{q}-1\right)\mathrm{d}\\ ⇒\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}2\mathrm{ap}-2\mathrm{aq}=\mathrm{q}\left(\mathrm{q}-1\right)\mathrm{d}-\mathrm{p}\left(\mathrm{p}-1\right)\mathrm{d}\\ ⇒\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}2\mathrm{a}\left(\mathrm{p}-\mathrm{q}\right)=\left({\mathrm{q}}^{2}-\mathrm{q}-{\mathrm{p}}^{2}+\mathrm{p}\right)\mathrm{d}\\ ⇒\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}2\mathrm{a}\left(\mathrm{p}-\mathrm{q}\right)=\left\{\mathrm{p}-\mathrm{q}-\left({\mathrm{p}}^{2}-{\mathrm{q}}^{2}\right)\mathrm{d}\right\}\\ ⇒\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}2\mathrm{a}\left(\mathrm{p}-\mathrm{q}\right)=\left\{\left(\mathrm{p}-\mathrm{q}\right)-\left(\mathrm{p}-\mathrm{q}\right)\left(\mathrm{p}+\mathrm{q}\right)\mathrm{d}\right\}\\ ⇒\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}2\mathrm{a}\left(\mathrm{p}-\mathrm{q}\right)=\left(\mathrm{p}-\mathrm{q}\right)\left\{1-\left(\mathrm{p}+\mathrm{q}\right)\mathrm{d}\right\}\\ ⇒\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}2\mathrm{a}=\left\{1-\left(\mathrm{p}+\mathrm{q}\right)\mathrm{d}\right\}\end{array}$ $\begin{array}{l}\mathrm{Now},{\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}S}}_{\text{p}+\text{q}}=\frac{\mathrm{p}+\mathrm{q}}{2}\left\{2\mathrm{a}+\left(\mathrm{p}+\mathrm{q}-1\right)\mathrm{d}\right\}\\ {\text{\hspace{0.17em}\hspace{0.17em} \hspace{0.17em} \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}S}}_{\text{p}+\text{q}}=\frac{\mathrm{p}+\mathrm{q}}{2}\left[\left\{1-\left(\mathrm{p}+\mathrm{q}\right)\right\}\mathrm{d}+\left(\mathrm{p}+\mathrm{q}-1\right)\mathrm{d}\right]\\ {\text{\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}S}}_{\text{p}+\text{q}}=\frac{\mathrm{p}+\mathrm{q}}{2}\left\{1-\left(\mathrm{p}+\mathrm{q}\right)+\left(\mathrm{p}+\mathrm{q}\right)-1\right\}\mathrm{d}\\ ⇒{\text{\hspace{0.17em} \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}S}}_{\text{p}+\text{q}}=0\\ \mathrm{Thus},\text{the sum of}\left(\mathrm{p}+\mathrm{q}\right)\text{terms is zero.}\end{array}$

Q.20

$\begin{array}{l}\text{Sum of the first p, q and r terms of an A.P. are a, b and c, respectively.}\\ \text{Prove that}\frac{\text{a}}{\text{p}}\left(\text{q}-\text{r}\right)\text{+}\frac{\text{b}}{\text{q}}\left(\text{r}-\text{p}\right)\text{+}\frac{\text{c}}{\text{r}}\left(\text{p}-\text{q}\right)\text{=0.}\end{array}$

Ans

$\begin{array}{l}\mathrm{Let}\text{first term of A.P. be A and common difference be d.}\\ \mathrm{Given}:\text{\hspace{0.17em}}{\mathrm{S}}_{\mathrm{p}}=\mathrm{a}⇒\mathrm{a}=\frac{\mathrm{p}}{2}\left\{2\mathrm{A}+\left(\mathrm{p}-1\right)\mathrm{d}\right\}\\ ⇒\frac{\mathrm{a}}{\mathrm{p}}=\frac{1}{2}\left\{2\mathrm{A}+\left(\mathrm{p}-1\right)\mathrm{d}\right\}\\ {\mathrm{S}}_{\mathrm{q}}=\mathrm{b}⇒\mathrm{b}=\frac{\mathrm{p}}{2}\left\{2\mathrm{A}+\left(\mathrm{q}-1\right)\mathrm{d}\right\}\\ ⇒\frac{\mathrm{b}}{\mathrm{q}}=\frac{1}{2}\left\{2\mathrm{A}+\left(\mathrm{q}-1\right)\mathrm{d}\right\}\\ \text{and}{\mathrm{S}}_{\mathrm{r}}=\mathrm{c}⇒\mathrm{c}=\frac{\mathrm{r}}{2}\left\{2\mathrm{A}+\left(\mathrm{r}-1\right)\mathrm{d}\right\}\\ ⇒\frac{\mathrm{c}}{\mathrm{r}}=\frac{1}{2}\left\{2\mathrm{A}+\left(\mathrm{r}-1\right)\mathrm{d}\right\}\end{array}$ $\begin{array}{l}\mathrm{L}.\mathrm{H}.\mathrm{S}.=\frac{\mathrm{a}}{\mathrm{p}}\left(\mathrm{q}-\mathrm{r}\right)+\frac{\mathrm{b}}{\mathrm{q}}\left(\mathrm{r}-\mathrm{p}\right)+\frac{\mathrm{c}}{\mathrm{r}}\left(\mathrm{p}-\mathrm{q}\right)\\ \text{\hspace{0.17em}\hspace{0.17em}}=\frac{1}{2}\left\{2\mathrm{A}+\left(\mathrm{p}-1\right)\mathrm{d}\right\}\left(\mathrm{q}-\mathrm{r}\right)+\frac{1}{2}\left\{2\mathrm{A}+\left(\mathrm{q}-1\right)\mathrm{d}\right\}\left(\mathrm{r}-\mathrm{p}\right)\\ +\frac{1}{2}\left\{2\mathrm{A}+\left(\mathrm{r}-1\right)\mathrm{d}\right\}\left(\mathrm{p}-\mathrm{q}\right)\\ \text{\hspace{0.17em}\hspace{0.17em}}=\frac{1}{2}\left[\begin{array}{l}2\mathrm{A}\left(\mathrm{q}-\mathrm{r}+\mathrm{r}-\mathrm{p}+\mathrm{p}-\mathrm{q}\right)+\\ \mathrm{d}\left\{\left(\mathrm{p}-1\right)\left(\mathrm{q}-\mathrm{r}\right)+\left(\mathrm{q}-1\right)\left(\mathrm{r}-\mathrm{p}\right)+\left(\mathrm{r}-1\right)\left(\mathrm{p}-\mathrm{q}\right)\right\}\end{array}\right]\\ \text{\hspace{0.17em}\hspace{0.17em}}=\frac{1}{2}\left\{2\mathrm{A}\left(0\right)+\mathrm{d}\left(\begin{array}{l}\mathrm{pq}-\mathrm{pr}-\mathrm{q}+\mathrm{r}+\mathrm{qr}-\mathrm{qp}-\mathrm{r}+\mathrm{p}\\ +\mathrm{rp}-\mathrm{rq}-\mathrm{p}+\mathrm{q}\end{array}\right)\right\}\\ \text{\hspace{0.17em}\hspace{0.17em}}=\frac{1}{2}\left(0+0\right)\\ \text{\hspace{0.17em}\hspace{0.17em}}=0=\mathrm{R}.\mathrm{H}.\mathrm{S}.\\ \mathrm{Hence}\text{proved.}\end{array}$

Q.21 The ratio of the sums of m and n terms of an A.P. is m2 : n2. Show that the ratio of mth and nth term is (2m – 1) : (2n – 1).

Ans

$\begin{array}{l}\mathrm{Let}\text{first term of an A.P.}=\text{a}\\ \text{and commong difference of A.P.}=\mathrm{d}\\ \mathrm{It}\text{is given that:}\\ {\text{S}}_{\text{m}}{\text{:S}}_{\text{n}}={\mathrm{m}}^{2}:{\mathrm{n}}^{2}\\ ⇒\text{\hspace{0.17em}}\frac{{\mathrm{S}}_{\mathrm{m}}}{{\mathrm{S}}_{\mathrm{n}}}=\frac{{\mathrm{m}}^{2}}{{\mathrm{n}}^{2}}\end{array}$ $\begin{array}{l}⇒\frac{\frac{\mathrm{m}}{2}\left\{2\mathrm{a}+\left(\mathrm{m}-1\right)\mathrm{d}\right\}}{\frac{\mathrm{n}}{2}\left\{2\mathrm{a}+\left(\mathrm{n}-1\right)\mathrm{d}\right\}}=\frac{{\mathrm{m}}^{2}}{{\mathrm{n}}^{2}}\\ ⇒\frac{\left\{2\mathrm{a}+\left(\mathrm{m}-1\right)\mathrm{d}\right\}}{\left\{2\mathrm{a}+\left(\mathrm{n}-1\right)\mathrm{d}\right\}}=\frac{\mathrm{m}}{\mathrm{n}}\\ ⇒2\mathrm{an}+\left(\mathrm{mn}-\mathrm{n}\right)\mathrm{d}=2\mathrm{am}+\left(\mathrm{mn}-\mathrm{m}\right)\mathrm{d}\\ ⇒2\mathrm{an}-2\mathrm{am}=\left(\mathrm{mn}-\mathrm{m}\right)\mathrm{d}-\left(\mathrm{mn}-\mathrm{n}\right)\mathrm{d}\\ ⇒\text{}2\mathrm{a}\left(\mathrm{n}-\mathrm{m}\right)=\mathrm{d}\left(\mathrm{mn}-\mathrm{m}-\mathrm{mn}+\mathrm{n}\right)\\ ⇒\text{}2\mathrm{a}\left(\mathrm{n}-\mathrm{m}\right)=\mathrm{d}\left(-\mathrm{m}+\mathrm{n}\right)\\ ⇒\text{}2\mathrm{a}=\mathrm{d}\\ \mathrm{Now},\\ \frac{{\mathrm{T}}_{\mathrm{m}}}{{\mathrm{T}}_{\mathrm{n}}}=\frac{\mathrm{a}+\left(\mathrm{m}-1\right)2\mathrm{a}}{\mathrm{a}+\left(\mathrm{n}-1\right)2\mathrm{a}}\\ =\frac{\mathrm{a}+\left(\mathrm{m}-1\right)2\mathrm{a}}{\mathrm{a}+\left(\mathrm{n}-1\right)2\mathrm{a}}\\ =\frac{1+2\mathrm{m}-2}{1+2\mathrm{n}-2}\\ =\frac{2\mathrm{m}-1}{2\mathrm{n}-1}\\ \mathrm{Thus},{\text{the ratio of m}}^{\text{th}}{\text{and n}}^{\text{th}}\text{term is}\left(2\mathrm{m}-1\right):\left(2\mathrm{n}-1\right).\end{array}$

Q.22 If the sum of n terms of an A.P. is 3n2 + 5n and its mth term is 164, find the value of m.

Ans

$\begin{array}{l}\mathrm{It}\text{is given that}\\ {\text{S}}_{\text{n}}={\text{3n}}^{\text{2}}+\text{5n}\\ \text{Replace n by n}-\text{1, we get}\\ {\text{S}}_{\text{n-1}}=\text{3}{\left(\text{n}-\text{1}\right)}^{\text{2}}+\text{5}\left(\text{n}-1\right)\\ \text{}={\text{3n}}^{\text{2}}-6\text{n}+\text{3}+\text{5n}-5\\ \text{}={\text{3n}}^{\text{2}}-\text{n}-2\\ \mathrm{So},{\text{T}}_{\text{n}}={\text{S}}_{\text{n}}-{\text{S}}_{\text{n-1}}\\ \text{}=\left({\text{3n}}^{\text{2}}+\text{5n}\right)-\left({\text{3n}}^{\text{2}}-\text{n}-2\right)\\ \text{}=6\text{n}+2\\ \mathrm{Since},{\text{T}}_{\text{m}}=164\\ \mathrm{So},\text{6m}+\text{2}=\text{164}\\ ⇒\text{m}=\frac{164-2}{6}\\ \text{}=\frac{162}{6}\\ \text{}=27\\ \mathrm{Thus},\text{the value of m is 27.}\end{array}$

Q.23 Insert five numbers between 8 and 26 such that the resulting sequence is an A.P.

Ans

$\begin{array}{l}\mathrm{Let}{\text{5 numbers between 8 and 26 are A}}_{\text{1}}{\text{, A}}_{\text{2}}{\text{, A}}_{\text{3}}{\text{, A}}_{\text{4}}{\text{and A}}_{\text{5}}\text{.}\\ {\text{Then, A.P. is 8, A}}_{\text{1}}{\text{, A}}_{\text{2}}{\text{, A}}_{\text{3}}{\text{, A}}_{\text{4}}{\text{, A}}_{\text{5}}\text{,\hspace{0.17em}26.}\\ âˆµ\mathrm{Last}\text{term}=\mathrm{a}+\left(\mathrm{n}-1\right)\mathrm{d}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}26=8+\left(7-1\right)\mathrm{d}\end{array}$ $\begin{array}{l}\text{\hspace{0.17em} \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}26=8+6\text{\hspace{0.17em}}\mathrm{d}\\ \text{\hspace{0.17em} \hspace{0.17em} \hspace{0.17em} \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{d}=\frac{26-8}{6}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\frac{18}{6}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=3\\ \mathrm{Then},\text{\hspace{0.17em}}{\mathrm{A}}_{1}=8+\mathrm{d}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=8+3\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=11\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}{\mathrm{A}}_{2}=8+2\mathrm{d}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=8+2\left(3\right)\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=14\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}{\mathrm{A}}_{3}=8+3\mathrm{d}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=8+3\left(3\right)\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=17\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}{\mathrm{A}}_{4}=8+4\mathrm{d}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=8+4\left(3\right)\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=20\end{array}$ $\begin{array}{l}\end{array}$ $\begin{array}{l}\text{\hspace{0.17em}\hspace{0.17em}}{\mathrm{A}}_{5}=8+5\mathrm{d}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=8+5\left(3\right)\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=23\\ \mathrm{Thus},\text{the 5 terms between 8 and 26 are 11, 14, 17, 20 and 23.}\end{array}$

Q.24

$\mathbf{\text{If}}\frac{{\mathbf{\text{a}}}^{\mathbf{\text{n}}}{\mathbf{\text{+b}}}^{\mathbf{\text{n}}}}{{\mathbf{\text{a}}}^{\mathbf{\text{n}}\mathbf{-}\mathbf{\text{1}}}{\mathbf{\text{+b}}}^{\mathbf{\text{n}}\mathbf{-}\mathbf{\text{1}}}}\mathbf{\text{\hspace{0.17em}\hspace{0.17em}is the A.M. between a and b, then find the value of n.}}$

Ans

$\begin{array}{l}\mathrm{A}.\mathrm{M}.\text{of a and b}=\frac{\mathrm{a}+\mathrm{b}}{2}\\ \text{\hspace{0.17em}}\frac{{\mathrm{a}}^{\mathrm{n}}+{\mathrm{b}}^{\mathrm{n}}}{{\mathrm{a}}^{\mathrm{n}–1}+{\mathrm{b}}^{\mathrm{n}–1}}=\frac{\mathrm{a}+\mathrm{b}}{2}\\ ⇒\left(\mathrm{a}+\mathrm{b}\right)\left({\mathrm{a}}^{\mathrm{n}–1}+{\mathrm{b}}^{\mathrm{n}–1}\right)\text{\hspace{0.17em}}=2\left({\mathrm{a}}^{\mathrm{n}}+{\mathrm{b}}^{\mathrm{n}}\right)\\ ⇒{\mathrm{a}}^{\mathrm{n}}+{\mathrm{ab}}^{\mathrm{n}-1}+{\mathrm{ba}}^{\mathrm{n}-1}+{\mathrm{b}}^{\mathrm{n}}=2{\mathrm{a}}^{\mathrm{n}}+2{\mathrm{b}}^{\mathrm{n}}\\ ⇒\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}{\mathrm{ab}}^{\mathrm{n}-1}-{\mathrm{b}}^{\mathrm{n}}={\mathrm{a}}^{\mathrm{n}}-{\mathrm{ba}}^{\mathrm{n}-1}\\ ⇒\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}{\mathrm{b}}^{\mathrm{n}-1}\left(\mathrm{a}-\mathrm{b}\right)=\text{\hspace{0.17em}}{\mathrm{a}}^{\mathrm{n}-1}\left(\mathrm{a}-\mathrm{b}\right)\\ ⇒\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}{\mathrm{b}}^{\mathrm{n}-1}=\text{\hspace{0.17em}}{\mathrm{a}}^{\mathrm{n}-1}\\ ⇒\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}{\left(\frac{\mathrm{a}}{\mathrm{b}}\right)}^{\mathrm{n}-1}=1=\text{\hspace{0.17em}}{\left(\frac{\mathrm{a}}{\mathrm{b}}\right)}^{0}\\ ⇒\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{n}=1\end{array}$

Q.25 Between 1 and 31, m numbers have been inserted in such a way that the resulting sequence is an A. P. and the ratio of 7th and (m – 1)th numbers is 5 : 9. Find the value of m.

Ans

$\begin{array}{l}\mathrm{Let}{\text{m numbers between 1 and 31 are A}}_{\text{1}}{\text{, A}}_{\text{2}}{\text{, A}}_{\text{3}}{\text{, A}}_{\text{4}}\text{,}...{\text{A}}_{\text{m-1}}{\text{,A}}_{\text{m}}\text{.}\\ {\text{Then, A.P. is 1, A}}_{\text{1}}{\text{, A}}_{\text{2}}{\text{, A}}_{\text{3}}{\text{, A}}_{\text{4}}\text{,}...{\text{A}}_{\text{m-1}}{\text{,A}}_{\text{m}}\text{,\hspace{0.17em}\hspace{0.17em}31.}\\ âˆµ\mathrm{Last}\text{term}=\mathrm{a}+\left(\mathrm{n}-1\right)\mathrm{d}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}31=1+\left(\mathrm{m}+2-1\right)\mathrm{d}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}30=\left(\mathrm{m}+1\right)\text{\hspace{0.17em}}\mathrm{d}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{d}=\frac{30}{\mathrm{m}+1}\\ \mathrm{Then},\text{\hspace{0.17em} \hspace{0.17em} \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}{\mathrm{A}}_{1}=1+\mathrm{d}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=1+\frac{30}{\mathrm{m}+1}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\frac{\mathrm{m}+31}{\mathrm{m}+1}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}{\mathrm{A}}_{7}=1+7\mathrm{d}\end{array}$ $\begin{array}{l}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=1+7\left(\frac{30}{\mathrm{m}+1}\right)\\ \text{\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}}=\frac{\mathrm{m}+211}{\mathrm{m}+1}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}{\mathrm{A}}_{\mathrm{m}-1}=1+\left(\mathrm{m}-1\right)\mathrm{d}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=1+\left(\mathrm{m}-1\right)\left(\frac{30}{\mathrm{m}+1}\right)\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\frac{\mathrm{m}+1+30\left(\mathrm{m}-1\right)}{\mathrm{m}+1}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\frac{31\mathrm{m}-29}{\mathrm{m}+1}\\ \mathrm{According}\text{to question:}\\ \frac{{\mathrm{A}}_{7}}{{\mathrm{A}}_{\mathrm{m}-1}}=\frac{\left(\frac{\mathrm{m}+211}{\mathrm{m}+1}\right)}{\left(\frac{31\text{\hspace{0.17em}}\mathrm{m}-29}{\mathrm{m}+1}\right)}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\frac{5}{9}=\frac{\text{\hspace{0.17em}}\mathrm{m}+211}{31\text{\hspace{0.17em}}\mathrm{m}-29}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}155\text{\hspace{0.17em}}\mathrm{m}-145=9\text{\hspace{0.17em}}\mathrm{m}+1899\\ \text{\hspace{0.17em} \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}155\text{\hspace{0.17em}}\mathrm{m}-9\mathrm{m}=1899+145\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}146\text{\hspace{0.17em}}\mathrm{m}=2044\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{m}=\frac{2044}{146}\\ =14\\ \mathrm{Thus},\text{the value of m is 14.}\end{array}$

Q.26 The difference between any two consecutive interior angles of a polygon is 5°. If the smallest angle is 120°, find the number of the sides of the polygon.

Ans

$\begin{array}{l}\mathrm{The}\text{difference between any two consecutive}\\ \text{angles}=5\mathrm{°}\\ \mathrm{The}\text{smallest angle of a polygon}=120°\\ \mathrm{Let}\text{number of sides in given polygon be n.Then,}\\ \text{Sum of angles of a polygon}=\left(\mathrm{n}-2\right)180\mathrm{°}\\ \mathrm{The}\text{series formed by angles of polygon is:}\\ \text{120°,125°,130°,135°,140°,\hspace{0.17em}}...\end{array}$ $\begin{array}{l}\mathrm{Sum}\text{of n angles of polygon of n sides}=\frac{\mathrm{n}}{2}\left\{2×120\mathrm{°}+\left(\mathrm{n}-1\right)×5\mathrm{°}\right\}\\ =\frac{\mathrm{n}}{2}\left\{240\mathrm{°}+5\mathrm{°}\text{\hspace{0.17em}}\mathrm{n}-5\mathrm{°}\right\}\\ =\frac{\mathrm{n}}{2}\left\{235\mathrm{°}+5\mathrm{°}\text{\hspace{0.17em}}\mathrm{n}\right\}\\ \mathrm{Therefore},\\ \frac{\mathrm{n}}{2}\left\{235\mathrm{°}+5\mathrm{°}\text{\hspace{0.17em}}\mathrm{n}\right\}=\left(\mathrm{n}-2\right)180\mathrm{°}\\ ⇒5\mathrm{°}\text{\hspace{0.17em}}{\mathrm{n}}^{2}+235\mathrm{°}\mathrm{n}=360\mathrm{°}\text{\hspace{0.17em}}\mathrm{n}-720\mathrm{°}\\ ⇒\text{\hspace{0.17em}}5\mathrm{°}\text{\hspace{0.17em}}{\mathrm{n}}^{2}-125\mathrm{°}\mathrm{n}+720\mathrm{°}=0\\ ⇒\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}{\mathrm{n}}^{2}-25\mathrm{°}\mathrm{n}+144\mathrm{°}=0\\ ⇒\text{\hspace{0.17em} \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\left(\mathrm{n}-9\right)\left(\mathrm{n}-16\right)=0\\ ⇒\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{n}=9,16\\ \mathrm{Number}\text{of sides in polygon is 9 because number of angles in}\\ \text{a polygon is equal to the number of sides. Sum of 16 angles}\\ \text{will be equal to sum of 9 angles if some angles are 0° or}\\ \text{negative,}\mathrm{which}\text{is impossible. Therefore number of sides in}\\ \text{the polygon is 9.}\end{array}$

Q.27

${\mathbf{\text{Find the 20}}}^{\mathbf{\text{th}}}{\mathbf{\text{and n}}}^{\mathbf{\text{th}}}\mathbf{\text{terms of the G.P.}}\frac{\mathbf{\text{5}}}{\mathbf{\text{2}}}\mathbf{\text{,}}\frac{\mathbf{\text{5}}}{\mathbf{\text{4}}}\mathbf{\text{,}}\frac{\mathbf{\text{5}}}{\mathbf{\text{8}}}\mathbf{\text{,}}\mathbf{.}\mathbf{..}$

Ans

$The\text{given G}\text{.P}\text{. is:}\text{}\text{}\text{}\frac{5}{2},\frac{5}{4},\frac{5}{8},\dots$ $\begin{array}{l}\mathrm{a}=\frac{5}{2}\\ \mathrm{r}=\frac{\left(\frac{5}{4}\right)}{\left(\frac{5}{2}\right)}=\frac{1}{2}\\ {\mathrm{T}}_{20}={\mathrm{ar}}^{20-1}\left[âˆµ{\mathrm{T}}_{\mathrm{n}}={\mathrm{ar}}^{\mathrm{n}-1}\right]\\ =\frac{5}{2}{\left(\frac{1}{2}\right)}^{19}\\ =\frac{5}{{2}^{20}}\\ \mathrm{and}\text{\hspace{0.17em}}{\mathrm{T}}_{\mathrm{n}}={\mathrm{ar}}^{\mathrm{n}-1}\\ =\frac{5}{2}{\left(\frac{1}{2}\right)}^{\mathrm{n}-1}\\ =\frac{5}{{2}^{\mathrm{n}}}\end{array}$

Q.28 Find the 12th term of a G.P. whose 8th term is 192 and the common ratio is 2.

Ans

$\begin{array}{l}\mathrm{Common}\text{ratio of given G.P.}=2\\ {8}^{\mathrm{th}}\text{term of A.P.}=192\\ \text{\hspace{0.17em}}{\mathrm{ar}}^{8-1}=192\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{a}.{2}^{7}=192\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{a}=\frac{192}{{2}^{7}}=1.5\\ {12}^{\mathrm{th}}\text{term of A.P.}={\mathrm{ar}}^{12-1}\end{array}$ $\begin{array}{l}=1.5{\left(2\right)}^{11}\\ =1.5×2048\\ =3072\end{array}$

Q.29 The 5th, 8th and 11th terms of a G.P. are p, q and s, respectively. Show that q2 = ps.

Ans

$\begin{array}{l}\text{Let first term of G.P. be a and common ratio be r.}\\ {\text{\hspace{0.17em}\hspace{0.17em}5}}^{\text{th}}\text{term of G.P.}=\mathrm{p}\\ ⇒\text{\hspace{0.17em}\hspace{0.17em}}{\mathrm{T}}_{5}=\mathrm{p}\\ ⇒\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{p}={\mathrm{ar}}^{4}\\ {\text{\hspace{0.17em}\hspace{0.17em}8}}^{\text{th}}\text{term of G.P.}=\mathrm{q}\\ ⇒\text{\hspace{0.17em}\hspace{0.17em}}{\mathrm{T}}_{8}=\mathrm{q}\\ ⇒\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{q}={\mathrm{ar}}^{7}\\ {\text{11}}^{\text{th}}\text{term of G.P.}=\mathrm{s}\\ ⇒{\mathrm{T}}_{11}=\mathrm{s}\\ ⇒\text{\hspace{0.17em} \hspace{0.17em}\hspace{0.17em}}\mathrm{s}={\mathrm{ar}}^{10}\\ \text{L.H.S.}={\mathrm{q}}^{2}\\ \text{\hspace{0.17em} \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}={\left({\mathrm{ar}}^{7}\right)}^{2}\\ \text{\hspace{0.17em} \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}={\mathrm{a}}^{2}{\mathrm{r}}^{14}\\ \text{R.H.S.=}\mathrm{ps}\\ \text{\hspace{0.17em} \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\left({\mathrm{ar}}^{4}\right)\left({\mathrm{ar}}^{10}\right)\\ \text{\hspace{0.17em} \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}={\mathrm{a}}^{2}{\mathrm{r}}^{14}\\ \text{So,\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}L.H.S.=R.H.S.}\\ \text{Hence proved.}\end{array}$

Q.30 The 4th term of a G.P. is square of its second term, and the first term is –3. Determine its 7th term.

Ans

$\begin{array}{l}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{First}\text{term of G.P.}=-3\\ \mathrm{Let}\text{common ratio of G.P.}=\mathrm{r}\\ \mathrm{According}\text{to given condition,}\\ {\mathrm{T}}_{4}={\left({\mathrm{T}}_{2}\right)}^{2}\\ ⇒\text{\hspace{0.17em}}{\mathrm{ar}}^{3}={\left(\mathrm{ar}\right)}^{2}\\ ⇒\text{\hspace{0.17em}\hspace{0.17em}}\left(-3\right){\mathrm{r}}^{3}={\left(-3\right)}^{2}{\mathrm{r}}^{2}\\ ⇒\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{r}=-3\\ \mathrm{Then},{\mathrm{T}}_{7}={\mathrm{ar}}^{6}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\left(-3\right){\left(-3\right)}^{6}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}={\left(-3\right)}^{7}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=-2187\\ \mathrm{Thus},{\text{7}}^{\text{th}}\text{term of G.P. is}-2187.\end{array}$

Q.31

$\begin{array}{l}\mathbf{\text{Which term of the following sequences:}}\\ \left(\text{a}\right)\mathbf{\text{. 2, 2}}\sqrt{\mathbf{\text{2}}}\mathbf{\text{, 4,}}\mathbf{.}\mathbf{..}\mathbf{\text{is 128?}}\\ \left(\text{b}\right)\mathbf{\text{. 3, 3}}\sqrt{\mathbf{\text{3}}}\mathbf{\text{, 3,}}\mathbf{.}\mathbf{..}\mathbf{\text{is 729?}}\\ \left(\text{c}\right)\mathbf{\text{. \hspace{0.17em}}}\frac{\mathbf{\text{1}}}{\mathbf{\text{3}}}\mathbf{\text{,\hspace{0.17em}}}\frac{\mathbf{\text{1}}}{\mathbf{\text{9}}}\mathbf{\text{,\hspace{0.17em}}}\frac{\mathbf{\text{1}}}{\mathbf{\text{27}}}\mathbf{\text{,}}\mathbf{.}\mathbf{..}\mathbf{\text{\hspace{0.17em} is \hspace{0.17em}}}\frac{\mathbf{\text{1}}}{\mathbf{\text{19683}}}\mathbf{\text{?}}\end{array}$

Ans

$\begin{array}{l}\left(\mathrm{a}\right)\mathrm{Let}{\text{t}}_{\text{n}}=128,\text{a}=2,\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} r}=\frac{2\sqrt{2}}{2}=\sqrt{2}\end{array}$ $\begin{array}{l}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}{\mathrm{t}}_{\mathrm{n}}={\mathrm{ar}}^{\mathrm{n}-1}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}128=2{\left(\sqrt{2}\right)}^{\mathrm{n}-1}\\ ⇒\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}{2}^{7}={\left(2\right)}^{1+\frac{\mathrm{n}-1}{2}}\\ ⇒\text{\hspace{0.17em} \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}7=1+\frac{\mathrm{n}-1}{2}\\ ⇒\text{\hspace{0.17em}}6×2=\mathrm{n}-1⇒\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{n}=13\\ \mathrm{Thus},{\text{128 is 13}}^{\text{th}}\text{term of G.P.}\\ \left(\mathrm{b}\right)\text{\hspace{0.17em}\hspace{0.17em}}\mathrm{Let}{\text{t}}_{\text{n}}=729,\text{a}=\sqrt{3},\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} r}=\frac{3}{\sqrt{3}}=\sqrt{3}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}{\mathrm{t}}_{\mathrm{n}}={\mathrm{ar}}^{\mathrm{n}-1}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}729=\sqrt{3}{\left(\sqrt{3}\right)}^{\mathrm{n}-1}\\ ⇒\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}{3}^{6}={\left(3\right)}^{\frac{1}{2}+\frac{\mathrm{n}-1}{2}}\\ ⇒\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}6=\frac{1}{2}+\frac{\mathrm{n}-1}{2}\\ ⇒\text{\hspace{0.17em}}6×2=1+\mathrm{n}-1\\ ⇒\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{n}=12\\ \mathrm{Thus},\text{}729{\text{is 12}}^{\text{th}}\text{term of G.P.}\\ \left(\mathrm{c}\right)\text{\hspace{0.17em}}\mathrm{Let}{\text{t}}_{\text{n}}=729,\text{a}=\frac{1}{3},\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}r}=\frac{\left(\frac{1}{9}\right)}{\left(\frac{1}{3}\right)}=\frac{1}{3}\end{array}$ $\begin{array}{l}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\end{array}$ $\begin{array}{l}{\mathrm{t}}_{\mathrm{n}}={\mathrm{ar}}^{\mathrm{n}-1}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\frac{1}{19683}=\frac{1}{3}{\left(\frac{1}{3}\right)}^{\mathrm{n}-1}\\ ⇒\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}{\left(\frac{1}{3}\right)}^{9}={\left(\frac{1}{3}\right)}^{\mathrm{n}}\\ ⇒\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}}9=\mathrm{n}\\ ⇒\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{n}=9\\ \mathrm{Thus},\text{}\frac{1}{19683}{\text{is 9}}^{\text{th}}\text{term of G.P.}\end{array}$

Q.32

$\mathbf{\text{For what values of x, the numbers\hspace{0.17em}\hspace{0.17em}}}\mathbf{-}\frac{\mathbf{\text{2}}}{\mathbf{\text{7}}}\mathbf{\text{,x,}}\mathbf{-}\frac{\mathbf{\text{7}}}{\mathbf{\text{2}}}\mathbf{\text{are in G.P.?}}$

Ans

$\begin{array}{l}-\frac{\text{2}}{\text{7}}\text{,x,}-\frac{\text{7}}{\text{2}}\text{are in G.P., then}\\ \text{Common ratio}=\frac{\mathrm{x}}{-\left(\frac{2}{7}\right)}\\ \text{\hspace{0.17em}\hspace{0.17em}}=-\frac{7\mathrm{x}}{2}\\ \text{Also, common ratio}=\frac{\left(-\frac{7}{2}\right)}{\mathrm{x}}\\ \text{\hspace{0.17em}\hspace{0.17em}}=-\frac{7}{2\mathrm{x}}\\ \text{Then,}\\ -\frac{7\mathrm{x}}{2}=-\frac{7}{2\mathrm{x}}\\ ⇒\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}{\mathrm{x}}^{2}=1\\ ⇒\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{x}=±1\end{array}$

Q.33

$\begin{array}{l}\mathbf{\text{Find the sum to indicated number of terms in each of the geometric}}\\ \mathbf{\text{progressions given below.}}\\ \mathbf{\text{i. 0.15, 0.015, 0.0015,… 20 terms.}}\\ \mathbf{\text{ii.}}\sqrt{\mathbf{\text{7}}}\mathbf{\text{,\hspace{0.17em}\hspace{0.17em}}}\sqrt{\mathbf{\text{21}}}\mathbf{\text{,\hspace{0.17em}\hspace{0.17em}3}}\sqrt{\mathbf{\text{7}}}\mathbf{\text{,\hspace{0.17em}\hspace{0.17em}}}\mathbf{.}\mathbf{..}\mathbf{\text{\hspace{0.17em}\hspace{0.17em}n\hspace{0.17em}terms.}}\\ \mathbf{\text{iii. 1,}}\mathbf{-}{\mathbf{\text{\hspace{0.17em}a, a}}}^{\mathbf{\text{2}}}\mathbf{\text{,}}\mathbf{-}{\mathbf{\text{\hspace{0.17em}a}}}^{\mathbf{\text{3}}}\mathbf{\text{,}}\mathbf{.}\mathbf{..}\mathbf{\text{n terms (if a}}\mathbf{\ne }\mathbf{-}\mathbf{\text{1).}}\\ {\mathbf{\text{iv. x}}}^{\mathbf{\text{3}}}{\mathbf{\text{, x}}}^{\mathbf{\text{5}}}{\mathbf{\text{, x}}}^{\mathbf{\text{7}}}\mathbf{\text{,}}\mathbf{.}\mathbf{..}\mathbf{\text{n terms (if x}}\mathbf{\ne }\mathbf{\text{± 1).}}\end{array}$

Ans

$\begin{array}{l}\mathrm{i}.\\ 0.\text{15},\text{}0.0\text{15},\text{}0.00\text{15},\dots \text{2}0\text{terms}\\ \mathrm{Since},\text{\hspace{0.17em}}{\mathrm{S}}_{\mathrm{n}}=\frac{\mathrm{a}\left(1-{\mathrm{r}}^{\mathrm{n}}\right)}{1-\mathrm{r}}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{a}=0.15,\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}r}=\frac{0.015}{0.15}=0.1\\ {\mathrm{S}}_{\mathrm{n}}=\frac{0.15\left\{1-{\left(0.1\right)}^{20}\right\}}{1-0.1}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\frac{0.15\left\{1-{\left(0.1\right)}^{20}\right\}}{0.9}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\frac{15}{90}\left\{1-{\left(0.1\right)}^{20}\right\}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\frac{1}{6}\left\{1-{\left(0.1\right)}^{20}\right\}\end{array}$ $\begin{array}{l}\mathrm{i}\mathrm{i}.\\ \sqrt{7},\text{\hspace{0.17em}}\sqrt{21},\text{\hspace{0.17em}\hspace{0.17em}}3\sqrt{7},\text{\hspace{0.17em}}...\text{\hspace{0.17em}}\mathrm{n}\text{\hspace{0.17em}}\mathrm{terms}\\ \mathrm{Since},\text{\hspace{0.17em}}{\mathrm{S}}_{\mathrm{n}}=\frac{\mathrm{a}\left({\mathrm{r}}^{\mathrm{n}}-1\right)}{\mathrm{r}-1}\end{array}$ $\begin{array}{l}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{}\end{array}$ $\begin{array}{l}\text{\hspace{0.17em}}\mathrm{a}=\sqrt{7},\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em} r}=\frac{\sqrt{21}}{\sqrt{7}}=\sqrt{3}\\ {\mathrm{S}}_{\mathrm{n}}=\frac{\sqrt{7}\left\{{\left(\sqrt{3}\right)}^{\mathrm{n}}-1\right\}}{\sqrt{3}-1}×\frac{\sqrt{3}+1}{\sqrt{3}+1}\\ \text{\hspace{0.17em} \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\frac{\sqrt{7}\left\{{\left(\sqrt{3}\right)}^{\mathrm{n}}-1\right\}}{3-1}×\left(\sqrt{3}+1\right)\end{array}$ $\begin{array}{l}\end{array}$ $\begin{array}{l}\mathrm{iii}.\\ 1,-\text{\hspace{0.17em}}\mathrm{a},{\mathrm{a}}^{2},-\text{\hspace{0.17em}}{\mathrm{a}}^{3},...\mathrm{nterms}\\ \mathrm{First}\text{term of G.P.}=1\\ \mathrm{Common}\text{ratio of G.P.}=\frac{-\mathrm{a}}{1}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=-\mathrm{a}\\ \mathrm{Since},\text{\hspace{0.17em}}{\mathrm{S}}_{\mathrm{n}}=\frac{\mathrm{a}\left(1-{\mathrm{r}}^{\mathrm{n}}\right)}{1-\mathrm{r}}\\ {\mathrm{S}}_{\mathrm{n}}=\frac{1\left\{1-{\left(-\mathrm{a}\right)}^{\mathrm{n}}\right\}}{1+\mathrm{a}}\end{array}$ $\begin{array}{l}\end{array}$ $\begin{array}{l}\mathrm{iv}.\\ {\mathrm{x}}^{3},{\mathrm{x}}^{5},{\mathrm{x}}^{7},...\mathrm{n}\mathrm{terms}\\ \mathrm{First}\text{term of G.P.}={\mathrm{x}}^{3}\end{array}$ $\begin{array}{l}\end{array}$ $\begin{array}{l}\mathrm{Common}\text{ratio of G.P.}=\frac{{\mathrm{x}}^{5}}{{\mathrm{x}}^{3}}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}={\mathrm{x}}^{2}\\ \mathrm{Since},\text{\hspace{0.17em}\hspace{0.17em}}{\mathrm{S}}_{\mathrm{n}}=\frac{\mathrm{a}\left({\mathrm{r}}^{\mathrm{n}}-1\right)}{\mathrm{r}-1}\\ \text{\hspace{0.17em}\hspace{0.17em}}{\mathrm{S}}_{\mathrm{n}}=\frac{{\mathrm{x}}^{3}\left\{1-{\left({\mathrm{x}}^{2}\right)}^{\mathrm{n}}\right\}}{1-{\mathrm{x}}^{2}}\\ {\mathrm{S}}_{\mathrm{n}}=\frac{{\mathrm{x}}^{3}\left(1-{\mathrm{x}}^{2\mathrm{n}}\right)}{1-{\mathrm{x}}^{2}}\end{array}$

Q.34

$\mathbf{Evaluate}\mathbf{:}\mathbf{\text{\hspace{0.17em}\hspace{0.17em}}}\sum _{k=1}^{11}\left(2+{3}^{k}\right).$

Ans

$\begin{array}{l}\mathrm{We}\mathrm{}\mathrm{have},\\ \sum _{\mathrm{k}=1}^{11}\left(2+{3}^{\mathrm{k}}\right)=\left(2+{3}^{1}\right)+\left(2+{3}^{2}\right)+\left(2+{3}^{3}\right)+...+\left(2+{3}^{11}\right)\\ \text{\hspace{0.17em}}=2×11+\left({3}^{1}+{3}^{2}+{3}^{3}+...+{3}^{11}\right)\\ \text{\hspace{0.17em}}=22+\frac{3\left({3}^{11}-1\right)}{3-1}\\ \text{\hspace{0.17em}}=22+\frac{3}{2}\left({3}^{11}-1\right)\end{array}$

Q.35

$\begin{array}{l}\mathbf{\text{The sum of first three terms of a G.P. is}}\frac{\mathbf{\text{39}}}{\mathbf{\text{10}}}\mathbf{\text{and their product is 1.}}\\ \mathbf{\text{Find the common ratio and the terms.}}\end{array}$

Ans

$\begin{array}{l}\mathrm{Let}\text{three terms in A.P. are}\frac{\mathrm{a}}{\mathrm{r}},\text{\hspace{0.17em}}\mathrm{a},\mathrm{ar}.\\ \mathrm{Then}\text{according to given conditions:}\\ \frac{\text{a}}{\mathrm{r}}+\mathrm{a}+\mathrm{ar}=\frac{39}{10}\\ \mathrm{a}\left(\frac{1}{\mathrm{r}}+1+\mathrm{r}\right)=\frac{39}{10}...\left(\mathrm{i}\right)\\ \mathrm{and}\text{\hspace{0.17em}}\frac{\mathrm{a}}{\mathrm{r}}×\text{\hspace{0.17em}}\mathrm{a}×\mathrm{ar}=1\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}{\mathrm{a}}^{3}=1\\ ⇒\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{a}=1\\ \mathrm{Substituting}\text{value of a in equation}\left(\mathrm{i}\right),\text{\hspace{0.17em}}\mathrm{we}\text{get}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}}1\left(\frac{1}{\mathrm{r}}+1+\mathrm{r}\right)=\frac{39}{10}\\ ⇒\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\left(\frac{1+\mathrm{r}+{\mathrm{r}}^{2}}{\mathrm{r}}\right)=\frac{39}{10}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}10\left({\mathrm{r}}^{2}+\mathrm{r}+1\right)=39\mathrm{r}\\ 10\text{\hspace{0.17em}}{\mathrm{r}}^{2}+10\text{\hspace{0.17em}}\mathrm{r}+10=39\mathrm{r}\\ 10\text{\hspace{0.17em}}{\mathrm{r}}^{2}-29\text{\hspace{0.17em}}\mathrm{r}+10=0\\ \text{\hspace{0.17em}}10\text{\hspace{0.17em}}{\mathrm{r}}^{2}-25\text{\hspace{0.17em}}\mathrm{r}-4\text{\hspace{0.17em}}\mathrm{r}+10=0\\ 5\mathrm{r}\left(2\mathrm{r}-5\right)-2\left(2\mathrm{r}-5\right)=0\\ \text{\hspace{0.17em}\hspace{0.17em}}\left(2\mathrm{r}-5\right)\left(2\mathrm{r}-5\right)=0\\ ⇒\mathrm{r}=\frac{5}{2},\frac{5}{2}\end{array}$ $\begin{array}{l}\mathrm{Thus},\text{first term of G.P. is 1 and common difference is}\frac{5}{2}.\\ \mathrm{Three}\text{terms are:\hspace{0.17em}}\frac{1}{\frac{5}{2}},1,1×\frac{5}{2}⇒\frac{2}{5},1,\frac{5}{2}\\ \mathrm{Or}\text{\hspace{0.17em}\hspace{0.17em}}\frac{1}{\frac{2}{5}},1,1×\frac{2}{5}⇒\frac{5}{2},1,\frac{2}{5}\text{\hspace{0.17em}\hspace{0.17em}}\end{array}$

Q.36 How many terms of G.P. 3, 32, 33, … are needed to give the sum 120?

Ans

$\begin{array}{l}\text{G}.\text{P}.\text{is 3},{\text{3}}^{\text{2}},{\text{3}}^{\text{3}},\text{}\dots \\ \mathrm{a}=3,\text{r}=\frac{{3}^{2}}{3}=3{\text{and S}}_{\text{n}}=120\\ {\mathrm{S}}_{\mathrm{n}}=\mathrm{a}\frac{\left({\mathrm{r}}^{\mathrm{n}}-1\right)}{\left(\mathrm{r}-1\right)}\end{array}$ $\begin{array}{l}120=3\left(\frac{{3}^{\mathrm{n}}-1}{3-1}\right)\\ 120=\frac{3}{2}\left({3}^{\mathrm{n}}-1\right)\\ 120×\frac{2}{3}=\left({3}^{\mathrm{n}}-1\right)\\ \text{\hspace{0.17em}}81={3}^{\mathrm{n}}\\ ⇒\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}{3}^{4}={3}^{\mathrm{n}}\\ ⇒\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{n}=4\\ \mathrm{Thus},\text{4 terms of G.P. are needed to give the sum 120.}\end{array}$

Q.37 The sum of first three terms of a G.P. is 16 and the sum of the next three terms is 128.
Determine the first term, the common ratio and the sum to n terms of the G.P.

Ans

$\begin{array}{l}\mathrm{Let}\text{first term of G.P. is a and common ratio is r. Then,}\\ {\text{six terms of G.P. are: a, ar, ar}}^{\text{2}}{\text{, ar}}^{\text{3}}{\text{, ar}}^{\text{4}}{\text{, ar}}^{\text{5}}{\text{, ar}}^{\text{6}}\text{.}\\ \text{According to first condition:}\\ \text{a}+\text{ar}+{\text{ar}}^{\text{2}}=16\text{\hspace{0.17em}}\\ ⇒\text{\hspace{0.17em}}\mathrm{a}\left(1+\mathrm{r}+{\mathrm{r}}^{2}\right)=16\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}...\left(\mathrm{i}\right)\\ {\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} ar}}^{\text{3}}+{\text{ar}}^{\text{4}}+{\text{ar}}^{\text{5}}=128\\ ⇒{\mathrm{ar}}^{3}\left(1+\mathrm{r}+{\mathrm{r}}^{2}\right)=128\text{\hspace{0.17em}\hspace{0.17em}}...\left(\mathrm{ii}\right)\\ \mathrm{Dividing}\text{equation}\left(\mathrm{ii}\right)\text{by equation}\left(\mathrm{i}\right),\text{we get}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\frac{{\mathrm{ar}}^{3}\left(1+\mathrm{r}+{\mathrm{r}}^{2}\right)}{\mathrm{a}\left(1+\mathrm{r}+{\mathrm{r}}^{2}\right)}=\frac{128}{16}\\ ⇒{\mathrm{r}}^{3}=8\\ ⇒\text{\hspace{0.17em}\hspace{0.17em}}\mathrm{r}=2\\ \mathrm{Putting}\text{r}=\text{2 in equation}\left(\mathrm{i}\right),\text{we get}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{a}\left(1+2+{2}^{2}\right)=16\\ \mathrm{a}=\frac{16}{7}\text{and r}=2\\ \mathrm{Sum}\text{of n terms is:}\\ {\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}S}}_{\text{n}}=\frac{\mathrm{a}\left({\mathrm{r}}^{\mathrm{n}}-1\right)}{\left(\mathrm{r}-1\right)}\end{array}$ $\begin{array}{l}\text{\hspace{0.17em}}=\frac{\frac{16}{7}\left({2}^{\mathrm{n}}-1\right)}{\left(2-1\right)}\\ \text{\hspace{0.17em}}=\frac{16}{7}\left({2}^{\mathrm{n}}-1\right)\end{array}$

Q.38 Given a G.P. with a = 729 and 7th term 64, determine S7.

Ans

$\begin{array}{l}\mathrm{First}\text{term of G.P.}=\left(\text{a}\right)\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=729\\ {7}^{\mathrm{th}}\text{}\mathrm{term}\text{of G.P.}={\mathrm{T}}_{7}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=64\\ \mathrm{Then},{\text{\hspace{0.17em}ar}}^{\text{6}}=64\\ ⇒\text{\hspace{0.17em}\hspace{0.17em}}729\text{\hspace{0.17em}}{\mathrm{r}}^{6}=64\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}{\mathrm{r}}^{6}=\frac{64}{729}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}{\mathrm{r}}^{6}=\frac{{2}^{6}}{{3}^{6}}\\ ⇒\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{r}=\frac{2}{3}\\ {\mathrm{S}}_{\mathrm{n}}=\frac{\mathrm{a}\left(1-{\mathrm{r}}^{\mathrm{n}}\right)}{\left(1-\mathrm{r}\right)}\\ ⇒\text{\hspace{0.17em}}{\mathrm{S}}_{7}=\frac{729\left(1-\frac{{2}^{\mathrm{n}}}{{3}^{\mathrm{n}}}\right)}{\left(1-\frac{2}{3}\right)}\end{array}$ $\begin{array}{l}⇒\text{\hspace{0.17em}\hspace{0.17em}}{\mathrm{S}}_{7}=729×3\left(1-\frac{{2}^{7}}{{3}^{7}}\right)\\ ⇒\text{\hspace{0.17em}\hspace{0.17em}}{\mathrm{S}}_{7}=729×3\left(1-\frac{128}{2187}\right)\\ ⇒\text{\hspace{0.17em}\hspace{0.17em}}{\mathrm{S}}_{7}=2187\left(\frac{2187-128}{2187}\right)\\ ⇒\text{\hspace{0.17em}\hspace{0.17em}}{\mathrm{S}}_{7}=2059\\ \mathrm{Thus},\text{sum of 7 terms of G.P. is 2059.}\end{array}$

Q.39 Find a G.P. for which sum of the first two terms is – 4 and the fifth term is 4 times the third term.

Ans

$\begin{array}{l}\mathrm{Let}\text{G.P. be:}\\ {\text{a, ar, ar}}^{\text{2}}{\text{, ar}}^{\text{3}}{\text{, ar}}^{\text{4}}{\text{, ar}}^{\text{5}}\text{,}...\\ \text{Then according to first condition,}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} a}+\mathrm{ar}=-\text{\hspace{0.17em}}4\\ \mathrm{or}\text{\hspace{0.17em}}\mathrm{a}\left(1+\mathrm{r}\right)=-4\\ \mathrm{or}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{a}=\frac{-4}{1+\mathrm{r}}...\left(\mathrm{i}\right)\\ \text{Then according to second condition,}\\ {\text{T}}_{\text{5}}={\text{4T}}_{\text{3}}\\ ⇒{\text{\hspace{0.17em}ar}}^{\text{4}}=4\left({\text{ar}}^{\text{2}}\right)\\ ⇒\text{\hspace{0.17em}\hspace{0.17em}}{\mathrm{r}}^{2}=4\\ ⇒\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{r}=±\text{\hspace{0.17em}}2\\ \mathrm{Putting}\text{r}=2\text{in equation}\left(\mathrm{i}\right),\text{we get}\end{array}$ $\begin{array}{l}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{a}=\frac{-4}{1+2}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=-\frac{4}{3}\\ \mathrm{Putting}\text{r}=2\text{in equation}\left(\mathrm{i}\right),\text{we get}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{a}=\frac{-4}{1-2}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\frac{-4}{-1}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=4\\ \mathrm{Therefore},\text{G.P. when a}=-\frac{4}{3}\text{and r}=\text{2}\\ -\frac{4}{3}\text{,}-\frac{4}{3}\left(2\right)\text{,}-\frac{4}{3}{\left(2\right)}^{\text{2}}\text{,}-\frac{4}{3}{\left(2\right)}^{\text{3}}\text{,}-\frac{4}{3}{\left(2\right)}^{\text{4}}\text{,}-\frac{4}{3}{\left(2\right)}^{\text{5}}\text{,}...\\ \text{or\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}-\frac{4}{3},-\frac{8}{3},-\frac{16}{3},-\frac{32}{3},-\frac{64}{3},-\frac{128}{3},...\\ \mathrm{Therefore},\text{G.P. when a}=4\text{and r}=-\text{2}\\ 4\text{,}4\left(-2\right)\text{,}4{\left(-2\right)}^{\text{2}}\text{,}4{\left(-2\right)}^{\text{3}}\text{,}4{\left(-2\right)}^{\text{4}}\text{,}4{\left(-2\right)}^{\text{5}}\text{,}...\\ \text{or\hspace{0.17em}\hspace{0.17em}}4,-8,16,-32,...\\ \mathrm{Therefore},\text{the required G.P. is}\\ \text{\hspace{0.17em}}-\frac{4}{3},-\frac{8}{3},-\frac{16}{3},-\frac{32}{3},-\frac{64}{3},-\frac{128}{3},...\\ \mathrm{or}\text{\hspace{0.17em}}4,-8,16,-32,...\end{array}$

Q.40 If the 4th, 10th and 16th terms of a G.P. are x, y and z, respectively. Prove that x, y, z are in G.P.

Ans

$\mathbf{Let}\mathbf{\text{first term of G.P. be ‘a’ and common ratio be ‘r’.}}$ $\begin{array}{l}{\text{\hspace{0.17em} T}}_{\text{4}}=\text{x}\\ ⇒\text{\hspace{0.17em}}{\mathrm{ar}}^{3}=\mathrm{x}\\ {\text{T}}_{\text{10}}=\text{y}\\ ⇒\text{\hspace{0.17em}}{\mathrm{ar}}^{9}=\mathrm{y}\\ {\text{T}}_{\text{16}}=\text{z}\\ ⇒\text{\hspace{0.17em}}{\mathrm{ar}}^{15}=\mathrm{z}\\ \mathrm{Now},\text{\hspace{0.17em}}\frac{\mathrm{y}}{\mathrm{x}}=\frac{{\mathrm{ar}}^{9}}{{\mathrm{ar}}^{3}}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}={\mathrm{r}}^{6}\\ \mathrm{and}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\frac{\mathrm{z}}{\mathrm{y}}=\frac{{\mathrm{ar}}^{15}}{{\mathrm{ar}}^{9}}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}={\mathrm{r}}^{6}\\ ⇒\frac{\mathrm{y}}{\mathrm{x}}=\frac{\mathrm{z}}{\mathrm{y}}\\ ⇒\mathrm{x},\mathrm{y},\mathrm{z}\text{are in G.P.}\end{array}$

Q.41 Find the sum to n terms of the sequence, 8, 88, 888, 8888… .

Ans

$\begin{array}{l}\mathrm{Let}\text{\hspace{0.17em}}{\mathrm{S}}_{\mathrm{n}}=\text{8}+\text{88}+\text{888}+\text{8888}+\dots \\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=8\left(1+11+111+1111+...\right)\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\frac{8}{9}\left(9+99+999+9999+...\right)\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\frac{8}{9}\left\{\left(10-1\right)+\left({10}^{2}-1\right)+\left({10}^{3}-1\right)+\left({10}^{4}-1\right)+...\right\}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\frac{8}{9}\left\{10+{10}^{2}+{10}^{3}+{10}^{4}+...-1-1-1-1-...\text{\hspace{0.17em}}\mathrm{n}\text{times}\right\}\end{array}$ $\begin{array}{l}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\frac{8}{9}\left\{\frac{10\left({10}^{\mathrm{n}}-1\right)}{10-1}-\text{\hspace{0.17em}}\mathrm{n}\right\}\left[âˆµ{\mathrm{S}}_{\mathrm{n}}=\frac{\mathrm{a}\left({\mathrm{r}}^{\mathrm{n}}-1\right)}{\mathrm{r}-1}\right]\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\frac{8}{9}\left\{\frac{10\left({10}^{\mathrm{n}}-1\right)}{9}-\text{\hspace{0.17em}}\mathrm{n}\right\}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\frac{80}{81}\left({10}^{\mathrm{n}}-1\right)-\frac{8}{9}\mathrm{n}\end{array}$

Q.42

$\begin{array}{l}\mathbf{\text{Find the sum of the products of the corresponding terms}}\\ \mathbf{\text{of the sequences 2, 4, 8,16, 32 and 128, 32, 8, 2,}}\frac{\mathbf{\text{1}}}{\mathbf{\text{2}}}\mathbf{\text{.}}\end{array}$

Ans

$\begin{array}{l}\text{Sum of the products of the corresponding terms of the sequences}\\ \text{2, 4, 8, 16, 32 and 128, 32, 8, 2,}\frac{1}{2}\\ =2×128+4×32+8×8+16×2+32×\frac{1}{2}\\ =256+128+64+32+16\\ =\frac{256\left\{1-{\left(\frac{1}{2}\right)}^{5}\right\}}{1-\frac{1}{2}}=\frac{256\left(\frac{32-1}{32}\right)}{\frac{1}{2}}\\ =256×2×\frac{31}{32}=496\end{array}$

Q.43 Show that the products of the corresponding terms of the sequences a, ar, ar2, … arn – 1 and A, AR, AR2, … ARn–1 form a G.P, and find the common ratio.

Ans

$\begin{array}{l}\text{The products of the corresponding terms of the sequences}\\ \text{a},\text{ar},{\text{ar}}^{\text{2}},\text{}\dots {{\text{ar}}^{\text{n}}}^{–\text{1}}\mathrm{}\text{and A},\text{AR},{\text{AR}}^{\text{2}},\text{}\dots {\text{AR}}^{\text{n}–\text{1}}\\ =\text{aA},\text{aARr},{\text{aAR}}^{2}{\text{r}}^{\text{2}},\text{}\dots {{\text{,aAR}}^{\text{n}}}^{–\text{1}}{{\text{r}}^{\text{n}}}^{–\text{1}}\\ \mathrm{Common}\text{ratio}=\frac{\text{aARr}}{\text{aA}}=\text{Rr}\\ \mathrm{Common}\text{ratio}=\frac{{\text{aAR}}^{2}{\text{r}}^{\text{2}}}{\text{aARr}}=\text{Rr}\\ \mathrm{Since},\text{common ratio is same i.e., rR.}\\ \text{So, aA},\text{aARr},{\text{aAR}}^{2}{\text{r}}^{\text{2}},\text{}\dots {{\text{,aAR}}^{\text{n}}}^{–\text{1}}{{\text{r}}^{\text{n}}}^{–\text{1}}\text{is in G.P.}\end{array}$

Q.44 Find four numbers forming a geometric progression in which the third term is greater than the first term by 9, and the second term is greater than the 4th by 18.

Ans

$\begin{array}{l}{\text{Let four numbers of G.P. are a, ar, ar}}^{\text{2}}{\text{, ar}}^{\text{3}}\text{.}\\ \text{According to first condition,}\\ {\text{\hspace{0.17em}\hspace{0.17em} T}}_{\text{3}}-{\text{T}}_{\text{1}}=9\\ ⇒\text{\hspace{0.17em}}{\mathrm{ar}}^{2}-\mathrm{a}=9\\ ⇒\text{\hspace{0.17em}}\mathrm{a}\left({\mathrm{r}}^{2}-1\right)=9...\left(\mathrm{i}\right)\\ \text{According to second condition,}\\ {\text{T}}_{\text{2}}-{\text{T}}_{\text{4}}=18\\ ⇒\text{\hspace{0.17em}}\mathrm{ar}-{\mathrm{ar}}^{3}=18\\ ⇒\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{ar}\left(1-{\mathrm{r}}^{2}\right)=18\\ ⇒\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{ar}\left({\mathrm{r}}^{2}-1\right)=-18...\left(\mathrm{ii}\right)\\ \text{Dividing equation}\left(\text{ii}\right)\text{by equation\hspace{0.17em}}\left(\text{i}\right)\text{, we get}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\frac{\mathrm{ar}\left({\mathrm{r}}^{2}-1\right)}{\mathrm{a}\left({\mathrm{r}}^{2}-1\right)}=-\frac{18}{9}\\ ⇒\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{r}=-2\\ \text{Putting r=}-\text{2, in equation}\left(\text{i}\right)\text{, we get}\\ \mathrm{a}\left\{{\left(-2\right)}^{2}-1\right\}=9\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{a}=\frac{9}{3}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=3\\ \text{Thus, the four terms of G.P. are:}\\ \text{3, 3}\left(-2\right)\text{, 3}{\left(-2\right)}^{\text{2}}\text{, 3}{\left(-2\right)}^{\text{3}}\\ \mathrm{i}.\mathrm{e}.,\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}3,-6,12,-24\end{array}$

Q.45

$\begin{array}{l}{\mathbf{\text{If the p}}}^{\mathbf{\text{th}}}{\mathbf{\text{, q}}}^{\mathbf{\text{th}}}{\mathbf{\text{\hspace{0.17em}and r}}}^{\mathbf{\text{th}}}\mathbf{\text{terms of a G.P. are a and b, respectively.}}\\ {\mathbf{\text{Prove that\hspace{0.17em} a}}}^{\mathbf{\text{q}}\mathbf{-}\mathbf{\text{r}}}{\mathbf{\text{\hspace{0.17em}b}}}^{\mathbf{\text{r}}\mathbf{-}\mathbf{\text{p}}}{\mathbf{\text{\hspace{0.17em}c}}}^{\mathbf{\text{p}}\mathbf{-}\mathbf{\text{q}}}\mathbf{\text{=1.}}\end{array}$

Ans

$\begin{array}{l}\mathrm{Let}\text{first term of G.P. is A and common ratio is R.}\\ \mathrm{Given}\text{that:}\\ {\text{T}}_{\text{p}}=\text{a}⇒{\mathrm{AR}}^{\mathrm{p}-1}=\mathrm{a}\\ {\text{T}}_{\text{q}}=\text{b}⇒{\mathrm{AR}}^{\mathrm{q}-1}=\mathrm{b}\\ {\text{T}}_{\text{r}}=\text{c}⇒{\mathrm{AR}}^{\mathrm{r}-1}=\mathrm{c}\\ \mathrm{L}.\mathrm{H}.\mathrm{S}.\text{\hspace{0.17em}}={\mathrm{a}}^{\mathrm{q}–\mathrm{r}}\text{\hspace{0.17em}}{\mathrm{b}}^{\mathrm{r}–\mathrm{p}}\text{\hspace{0.17em}}{\mathrm{c}}^{\mathrm{p}–\mathrm{q}}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}}={\left({\mathrm{AR}}^{\mathrm{p}-1}\right)}^{\mathrm{q}–\mathrm{r}}\text{\hspace{0.17em}}{\left({\mathrm{AR}}^{\mathrm{q}-1}\right)}^{\mathrm{r}–\mathrm{p}}\text{\hspace{0.17em}}{\left({\mathrm{AR}}^{\mathrm{r}-1}\right)}^{\mathrm{p}–\mathrm{q}}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}={\mathrm{A}}^{\mathrm{q}-\mathrm{r}+\mathrm{r}-\mathrm{p}+\mathrm{p}-\mathrm{q}}\text{\hspace{0.17em}}{\mathrm{R}}^{\left(\mathrm{p}-1\right)\left(\mathrm{q}-\mathrm{r}\right)+\left(\mathrm{q}-1\right)\left(\mathrm{r}-\mathrm{p}\right)\left(\mathrm{r}-1\right)\left(\mathrm{p}-\mathrm{q}\right)}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}={\mathrm{A}}^{0}{\mathrm{R}}^{\mathrm{pq}-\mathrm{pr}-\mathrm{q}+\mathrm{r}+\mathrm{qr}-\mathrm{qp}-\mathrm{r}+\mathrm{p}+\mathrm{rp}-\mathrm{rq}-\mathrm{p}+\mathrm{q}}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}={\mathrm{A}}^{0}{\mathrm{R}}^{0}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=1×1\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=1=\mathrm{R}.\mathrm{H}.\mathrm{S}.\end{array}$

Q.46 If the first and the nth term of a G.P. are a and b, respectively, and if P is the product of n terms, prove that P2 = (ab)n.

Ans

$\begin{array}{l}\mathrm{First}\text{term of G.P.}=\text{a}\\ {\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}n}}^{\text{th}}\text{term of G.P.}=\text{b}\\ \text{Let common ratio}=\mathrm{r}\\ \mathrm{The}\text{G.P. is}\\ {\text{a, ar, ar}}^{\text{2}}{\text{, ar}}^{\text{3}}{\text{, ar}}^{\text{4}}\text{,}...\text{, b}\\ \text{and b}={\text{ar}}^{\text{n-1}}\\ \text{Product of n terms}=\text{a}×\text{ar}×{\text{ar}}^{\text{2}}×{\text{ar}}^{\text{3}}×{\text{ar}}^{\text{4}}×...×\text{b}\\ \text{\hspace{0.17em}\hspace{0.17em}}=\text{a}×\text{ar}×{\text{ar}}^{\text{2}}×{\text{ar}}^{\text{3}}×{\text{ar}}^{\text{4}}×...×{\text{ar}}^{\text{n-1}}\end{array}$ $\begin{array}{l}\text{\hspace{0.17em}\hspace{0.17em}}={\mathrm{a}}^{\mathrm{n}}×{\mathrm{r}}^{1+2+3+4+...+\left(\mathrm{n}-1\right)}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{P}\text{\hspace{0.17em}}={\mathrm{a}}^{\mathrm{n}}×{\mathrm{r}}^{\frac{\left(\mathrm{n}-1\right)\left(\mathrm{n}-1+1\right)}{2}}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{P}\text{\hspace{0.17em}}={\mathrm{a}}^{\mathrm{n}}×{\mathrm{r}}^{\frac{\mathrm{n}\left(\mathrm{n}-1\right)}{2}}\\ \mathrm{Squarring}\text{both sides, we get}\\ {\text{P}}^{\text{2}}={\mathrm{a}}^{2\mathrm{n}}{\mathrm{r}}^{\mathrm{n}\left(\mathrm{n}-1\right)}\\ ={\left({\mathrm{a}}^{2}{\mathrm{r}}^{\mathrm{n}-1}\right)}^{\mathrm{n}}\\ ={\left(\mathrm{a}×{\mathrm{ar}}^{\mathrm{n}-1}\right)}^{\mathrm{n}}\\ {\text{\hspace{0.17em}P}}^{\text{2}}={\left(\mathrm{ab}\right)}^{\mathrm{n}}\left[âˆµ\mathrm{b}={\mathrm{ar}}^{\mathrm{n}-1}\right]\\ \mathrm{Hence}\text{Proved.}\end{array}$

Q.47

$\begin{array}{l}\mathbf{\text{Show that the ratio of the sum of first n terms of a G.P.}}\\ \mathbf{\text{to the sum of terms from \hspace{0.17em}}}{\left(\text{n+1}\right)}^{\mathbf{\text{th}}}\mathbf{\text{to}}{\left(\text{2n}\right)}^{\mathbf{\text{th}}}\mathbf{\text{term is}}\frac{\mathbf{\text{1}}}{{\mathbf{\text{r}}}^{\mathbf{\text{n}}}}\mathbf{.}\end{array}$

Ans

$\begin{array}{l}\text{LetfirsttermofG.P.isaandcommonratioisr.Then,}\\ \text{Sumoffirstnterms}\left({\text{S}}_{\text{n}}\right)\text{=}\frac{\text{a}\left({\text{r}}^{\text{n}}\text{-1}\right)}{\left(\text{r-1}\right)}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}Sumof2nterms}\left({\text{S}}_{\text{2n}}\right)\text{=}\frac{\text{a}\left({\text{r}}^{\text{2n}}\text{-1}\right)}{\left(\text{r-1}\right)}\\ \text{Sumoftermsfrom\hspace{0.17em}}{\left(\text{n+1}\right)}^{\text{th}}\text{to}{\left(\text{2n}\right)}^{\text{th}}\text{term}\\ {\text{=S}}_{\text{2n}}{\text{-S}}_{\text{n}}\end{array}$ $\begin{array}{l}\text{=}\frac{\text{a}\left({\text{r}}^{\text{2n}}\text{-1}\right)}{\left(\text{r-1}\right)}\text{–}\frac{\text{a}\left({\text{r}}^{\text{n}}\text{-1}\right)}{\left(\text{r-1}\right)}\\ \text{=}\frac{\text{a}}{\left(\text{r-1}\right)}\left({\text{r}}^{\text{2n}}{\text{-1-r}}^{\text{n}}\text{+1}\right)\\ \text{=}\frac{{\text{ar}}^{\text{n}}}{\left(\text{r-1}\right)}\left({\text{r}}^{\text{n}}\text{-1}\right)\\ \text{The required ratio=}\frac{{\text{S}}_{\text{n}}}{{\text{S}}_{\text{2n}}{\text{-S}}_{\text{n}}}\\ \text{=}\frac{\frac{\text{a}\left({\text{r}}^{\text{n}}\text{-1}\right)}{\left(\text{r-1}\right)}}{\frac{{\text{ar}}^{\text{n}}}{\left(\text{r-1}\right)}\left({\text{r}}^{\text{n}}\text{-1}\right)}\\ \text{=}\frac{\text{1}}{{\text{r}}^{\text{n}}}\\ \text{Thus,\hspace{0.17em}\hspace{0.17em}the ratio of the sum of first n terms of a G.P. to}\\ \text{the sum of terms from \hspace{0.17em}}{\left(\text{n+1}\right)}^{\text{th}}\text{to}{\left(\text{2n}\right)}^{\text{th}}\text{term is}\frac{\text{1}}{{\text{r}}^{\text{n}}\text{.}}\end{array}$

Q.48 If a, b, c and d are in G.P. show that (a2 + b2 + c2) (b2 + c2 + d2) = (ab + bc + cd)2.

Ans

$\begin{array}{l}\mathrm{Since},\text{a, b, c, d are in G.P.}\\ \text{So,}\frac{\mathrm{b}}{\mathrm{a}}=\frac{\mathrm{c}}{\mathrm{b}}=\frac{\mathrm{d}}{\mathrm{c}}=\mathrm{r}\left(\mathrm{let}\right)\\ ⇒\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{b}=\mathrm{ar},\text{c}=\mathrm{br}={\mathrm{ar}}^{2}\\ \mathrm{and}\text{d}=\mathrm{cr}={\mathrm{ar}}^{4}\\ \mathrm{L}.\mathrm{H}.\mathrm{S}.=\left({\text{a}}^{\text{2}}+{\text{b}}^{\text{2}}+{\text{c}}^{\text{2}}\right)\left({\text{b}}^{\text{2}}+{\text{c}}^{\text{2}}+{\text{d}}^{\text{2}}\right)\end{array}$ $\begin{array}{l}\text{\hspace{0.17em}\hspace{0.17em}}=\left({\text{a}}^{\text{2}}+{\text{a}}^{\text{2}}{\mathrm{r}}^{2}+{\text{a}}^{\text{2}}{\mathrm{r}}^{4}\right)\left({\text{a}}^{\text{2}}{\mathrm{r}}^{2}+{\text{a}}^{\text{2}}{\mathrm{r}}^{4}+{\text{a}}^{\text{2}}{\mathrm{r}}^{6}\right)\\ \text{\hspace{0.17em}\hspace{0.17em}}={\mathrm{a}}^{2}\left(1+{\mathrm{r}}^{2}+{\mathrm{r}}^{4}\right)×{\text{a}}^{\text{2}}{\mathrm{r}}^{2}\left(1+{\mathrm{r}}^{2}+{\mathrm{r}}^{4}\right)\\ \text{\hspace{0.17em}\hspace{0.17em}}={\mathrm{a}}^{4}{\mathrm{r}}^{2}{\left(1+{\mathrm{r}}^{2}+{\mathrm{r}}^{4}\right)}^{2}\\ \mathrm{R}.\mathrm{H}.\mathrm{S}.={\left(\text{ab}+\text{bc}+\text{cd}\right)}^{\text{2}}\\ \text{\hspace{0.17em}\hspace{0.17em}}={\left(\text{a}×\mathrm{ar}\text{}+\text{}\mathrm{ar}×{\mathrm{ar}}^{2}\text{}+\text{}{\mathrm{ar}}^{2}×{\mathrm{ar}}^{4}\right)}^{\text{2}}\\ \text{\hspace{0.17em}\hspace{0.17em}}={\mathrm{a}}^{4}{\mathrm{r}}^{2}{\left(1+{\mathrm{r}}^{2}+{\mathrm{r}}^{4}\right)}^{2}\\ \mathrm{L}.\mathrm{H}.\mathrm{S}.=\mathrm{R}.\mathrm{H}.\mathrm{S}.\\ \mathrm{Therefore},\\ \left({\text{a}}^{\text{2}}+{\text{b}}^{\text{2}}+{\text{c}}^{\text{2}}\right)\left({\text{b}}^{\text{2}}+{\text{c}}^{\text{2}}+{\text{d}}^{\text{2}}\right)={\left(\text{ab}+\text{bc}+\text{cd}\right)}^{\text{2}}\\ \mathrm{Hence}\text{proved.}\end{array}$

Q.49 Insert two numbers between 3 and 81 so that the resulting sequence is G.P.

Ans

$\begin{array}{l}\mathrm{Let}\text{two numbers between 3 and 81 are x and y in such a way}\\ \text{that 3, x, y, 81 are in G.P.}\\ \text{So, first term of G.P.}=\text{3}\\ \text{\hspace{0.17em}}\mathrm{Common}\text{difference}=\mathrm{r}\\ \text{\hspace{0.17em}\hspace{0.17em}}{4}^{\mathrm{th}}\text{\hspace{0.17em}}\mathrm{term}\text{of G.P.}={\mathrm{ar}}^{3}\\ ⇒\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}3\text{\hspace{0.17em}}{\mathrm{r}}^{3}=81\\ ⇒\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}\hspace{0.17em}}{\mathrm{r}}^{3}=\frac{81}{3}\\ ⇒\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{r}=\sqrt{27}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=3\end{array}$ $\begin{array}{l}\therefore \mathrm{x}={\mathrm{T}}_{2}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}}=\mathrm{ar}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=3×3\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}}=9\\ \mathrm{andy}={\mathrm{T}}_{3}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}={\mathrm{ar}}^{2}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=3×{3}^{2}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=27\\ \mathrm{Thus},\text{two numbers between 3 and 81 are 9 and 27.}\end{array}$

Q.50

$\mathbf{\text{Find the value of n so that\hspace{0.17em}}}\frac{{\mathbf{\text{a}}}^{\mathbf{\text{n+1}}}{\mathbf{\text{+b}}}^{\mathbf{\text{n+1}}}}{{\mathbf{\text{a}}}^{\mathbf{\text{n}}}{\mathbf{\text{+b}}}^{\mathbf{\text{n}}}}\mathbf{\text{may be the geometric mean between a and b.}}$

Ans

$\begin{array}{l}\mathrm{Geometric}\text{mean between a and b}=\sqrt{\mathrm{ab}}\\ \frac{{\text{a}}^{\text{n}+1}+{\text{b}}^{\text{n}+1}}{{\text{a}}^{\text{n}}+{\text{b}}^{\text{n}}}=\sqrt{\mathrm{ab}}\\ ⇒\text{}{\left(\mathrm{ab}\right)}^{\frac{1}{2}}\left({\text{a}}^{\text{n}}+{\text{b}}^{\text{n}}\right)={\text{a}}^{\text{n}+1}+{\text{b}}^{\text{n}+1}\\ ⇒{\text{a}}^{\text{n}+\frac{1}{2}}{\text{b}}^{\frac{1}{2}}+{\text{a}}^{\frac{1}{2}}{\text{b}}^{\text{n}+\frac{1}{2}}={\text{a}}^{\text{n}+1}+{\text{b}}^{\text{n}+1}\\ ⇒{\text{a}}^{\text{n}+\frac{1}{2}}{\text{b}}^{\frac{1}{2}}-{\text{a}}^{\text{n}+1}={\text{b}}^{\text{n}+1}-{\text{a}}^{\frac{1}{2}}{\text{b}}^{\text{n}+\frac{1}{2}}\\ ⇒{\text{a}}^{\text{n}+\frac{1}{2}}\left({\text{b}}^{\frac{1}{2}}-{\text{a}}^{\frac{1}{2}}\right)={\text{b}}^{\text{n}+\frac{1}{2}}\left({\text{b}}^{\frac{1}{2}}-{\text{a}}^{\frac{1}{2}}\right)\\ ⇒{\text{a}}^{\text{n}+\frac{1}{2}}={\text{b}}^{\text{n}+\frac{1}{2}}\\ ⇒\text{}{\left(\frac{\text{a}}{\text{b}}\right)}^{\text{n}+\frac{1}{2}}=1\\ ⇒\text{}{\left(\frac{\text{a}}{\text{b}}\right)}^{\text{n}+\frac{1}{2}}={\left(\frac{\text{b}}{\text{a}}\right)}^{0}\\ ⇒\text{n}+\frac{1}{2}=0\\ ⇒\text{n}=-\frac{1}{2}\end{array}$

Q.51

$\begin{array}{l}\mathbf{\text{The sum of two numbers is 6 times their geometric mean, show that}}\\ \mathbf{\text{numbers are in the ratio}}\left(\text{3+2}\sqrt{\text{2}}\right)\mathbf{\text{:}}\left(\text{3}-\text{2}\sqrt{\text{2}}\right)\mathbf{\text{.}}\end{array}$

Ans

$\begin{array}{l}\mathrm{Let}\text{two numbers be a and b.}\\ \text{G.M. of a and b}=\sqrt{\mathrm{ab}}\\ \mathrm{Then},\text{according to given condition,}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}a}+\text{b}=6\sqrt{\mathrm{ab}}...\left(\mathrm{i}\right)\\ âˆµ\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}{\left(\text{a}-\text{b}\right)}^{2}={\left(\text{a}+\text{b}\right)}^{2}-4\mathrm{ab}\\ ⇒\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}{\left(\text{a}-\text{b}\right)}^{2}={\left(6\sqrt{\mathrm{ab}}\right)}^{2}-4\mathrm{ab}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=36\mathrm{ab}-4\mathrm{ab}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=32\mathrm{ab}\\ ⇒\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} a}-\text{b}=\sqrt{32\mathrm{ab}}\\ ⇒\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} a}-\text{b}=4\sqrt{2}\sqrt{\mathrm{ab}}...\left(\mathrm{ii}\right)\end{array}$ $\begin{array}{l}\mathrm{Solving}\text{equation}\left(\mathrm{i}\right)\text{and equation}\left(\mathrm{ii}\right),\text{we get}\\ \text{a}=\left(3+2\sqrt{2}\right)\sqrt{\mathrm{ab}}\text{and b}=\left(3-2\sqrt{2}\right)\sqrt{\mathrm{ab}}\\ \mathrm{Then},\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}ratio of the numbers}=\frac{\left(3+2\sqrt{2}\right)\sqrt{\mathrm{ab}}}{\left(3-2\sqrt{2}\right)\sqrt{\mathrm{ab}}}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\left(3+2\sqrt{2}\right):\left(3-2\sqrt{2}\right)\end{array}$

Q.52

$\begin{array}{l}\mathbf{\text{If A and G be A.M. and G.M., respectively between two positive numbers,}}\\ \mathbf{\text{prove that the numbers are\hspace{0.17em}\hspace{0.17em}A±}}\sqrt{\left(\text{A+G}\right)\left(\text{A}-\text{G}\right)}\mathbf{\text{.}}\end{array}$

Ans

$\begin{array}{l}\text{Let two numbers are a and b.}\\ \text{Then, A.M.}=\frac{\mathrm{a}+\mathrm{b}}{2}\\ ⇒\mathrm{A}=\frac{\mathrm{a}+\mathrm{b}}{2}\\ ⇒\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{a}+\mathrm{b}=2\mathrm{A}...\left(\mathrm{i}\right)\\ âˆµ\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{G}.\mathrm{M}.=\sqrt{\mathrm{ab}}\\ ⇒\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{G}=\sqrt{\mathrm{ab}}\\ ⇒\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}}{\mathrm{G}}^{2}=\mathrm{ab}...\left(\mathrm{ii}\right)\\ âˆµ\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{a}-\mathrm{b}=\sqrt{{\left(\mathrm{a}+\mathrm{b}\right)}^{2}-4\mathrm{ab}}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\sqrt{{\left(2\mathrm{A}\right)}^{2}-4{\mathrm{G}}^{2}}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{a}-\mathrm{b}=2\sqrt{{\mathrm{A}}^{2}-{\mathrm{G}}^{2}}...\left(\mathrm{iii}\right)\\ \text{Solving equation}\left(\text{i}\right)\text{and equation}\left(\text{iii}\right)\text{, we get}\\ \text{a}=\mathrm{A}+\sqrt{{\mathrm{A}}^{2}-{\mathrm{G}}^{2}}\text{and b}=\mathrm{A}-\sqrt{{\mathrm{A}}^{2}-{\mathrm{G}}^{2}}\\ \text{Therefore, the two numbers are:}\\ \mathrm{A}±\sqrt{\left(\mathrm{A}+\mathrm{G}\right)\left(\mathrm{A}-\mathrm{G}\right)}.\end{array}$

Q.53 The number of bacteria in a certain culture doubles every hour. If there were 30 bacteria present in the culture originally, how many bacteria will be present at the end of 2nd hour, 4th hour and nth hour?

Ans

$\begin{array}{l}\mathrm{Originally}\text{present bacteria in a culture}=30\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{Number}\text{of bacteria after one hour}=60\\ \mathrm{Since},\text{number of bacteria doubles every hour.}\\ \text{So, common ratio}\left(\mathrm{r}\right)\text{in G.P.}=2\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{First}\text{term}\left(\mathrm{a}\right)\text{in G.P.}=30\\ \mathrm{Number}{\text{of bacteria after 2}}^{\text{nd}}\text{hour}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}={\text{ar}}^{\text{2}}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=30×{2}^{2}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=120\\ \mathrm{Number}{\text{of bacteria after 4}}^{\text{th}}\text{hour}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}={\text{ar}}^{\text{4}}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=30×{2}^{4}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=480\end{array}$ $\begin{array}{l}\mathrm{Number}{\text{of bacteria after n}}^{\text{th}}\text{hour}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}={\text{ar}}^{\text{n}}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=30×{2}^{\mathrm{n}}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=30\left({2}^{\mathrm{n}}\right)\end{array}$

Q.54 What will â‚¹ 500 amounts to in 10 years after its deposit in a bank which pays annual interest rate of 10% compounded annually?

Ans

$\begin{array}{l}\text{Principal amount deposited in the bank}=\text{\hspace{0.17em}}â‚¹\text{\hspace{0.17em}}500\\ \mathrm{Rate}\text{of interest in the bank}=10\mathrm{%}\text{\hspace{0.17em}}\mathrm{p}.\mathrm{a}.\\ \mathrm{Number}\text{of years for amount deposited}=10\text{years}\\ \mathrm{Amount}\text{after one year in the bank}=500\left(1+\frac{10}{100}\right)\\ =500\left(\frac{11}{10}\right)\\ \mathrm{Amount}\text{after two years in the bank}=500{\left(1+\frac{10}{100}\right)}^{2}\\ =500{\left(\frac{11}{10}\right)}^{2}\\ \mathrm{Amount}\text{after three years in the bank}=500{\left(1+\frac{10}{100}\right)}^{3}\\ =500{\left(\frac{11}{10}\right)}^{3}\\ \mathrm{So},\text{G.P. formed by using amount obtained from bank is}\end{array}$ $\begin{array}{l}500\left(\frac{11}{10}\right),500{\left(\frac{11}{10}\right)}^{2},500{\left(\frac{11}{10}\right)}^{3},...\\ \mathrm{First}\text{term of G.P.}=500\left(\frac{11}{10}\right)\\ \mathrm{Common}\text{ratio of G.P.}=\frac{11}{10}\\ \mathrm{Number}\text{of years}=10\\ \mathrm{Since},{\text{T}}_{\mathrm{n}}={\mathrm{ar}}^{\mathrm{n}-1}\\ ⇒{\text{T}}_{10}=500\left(\frac{11}{10}\right)×{\left(\frac{11}{10}\right)}^{10-1}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}}=500{\left(1.1\right)}^{10}\\ \mathrm{Thus},\text{}â‚¹\text{\hspace{0.17em}500 amounts to}â‚¹\text{\hspace{0.17em}}500{\left(1.1\right)}^{10}\text{in 10 years after its}\\ \text{deposit in a bank.}\end{array}$

Q.55 If A.M. and G.M. of roots of a quadratic equation are 8 and 5, respectively, then obtain the quadratic equation.

Ans

$\begin{array}{l}\mathrm{Let}\text{roots of a quadratic equation be a and b.}\\ \text{A.M. of a and b}=8\\ ⇒\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\frac{\mathrm{a}+\mathrm{b}}{2}=8\\ ⇒\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{a}+\mathrm{b}=16...\left(\mathrm{i}\right)\\ \text{\hspace{0.17em}\hspace{0.17em}}\mathrm{G}.\mathrm{M}.\text{of a and b}=5\\ ⇒\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\sqrt{\mathrm{ab}}=5\\ ⇒\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}}\mathrm{ab}=25...\left(\mathrm{ii}\right)\\ \mathrm{Quadratic}\text{equation, whose roots are a and b,}\end{array}$ $\begin{array}{l}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}}\left(\mathrm{x}-\mathrm{a}\right)\left(\mathrm{x}-\mathrm{b}\right)=0\\ {\mathrm{x}}^{2}-\left(\mathrm{a}+\mathrm{b}\right)\mathrm{x}+\mathrm{ab}=0\\ \text{\hspace{0.17em} \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}{\mathrm{x}}^{2}-16\mathrm{x}+25=0\left[\mathrm{From}\text{equation}\left(\mathrm{i}\right)\text{and}\left(\mathrm{ii}\right)\right]\\ \mathrm{Thus},\text{the required quadratic equation is \hspace{0.17em}}{\mathrm{x}}^{2}-16\mathrm{x}+25=0.\end{array}$

Q.56 Find the sum to n terms of each of the series in Exercises 1 to 7.

1. 1 × 2 + 2 × 3 + 3 × 4 + 4 × 5 +…
2. 1 × 2 × 3 + 2 × 3 × 4 + 3 × 4 × 5 + …
3. 3 × 12 + 5 × 22 + 7 × 32 + …
4.

$\frac{1}{1×2}+\frac{1}{2×3}+\frac{1}{3×4}+\dots$

5. 52 + 62 + 72 + … + 202
6. 3 × 8 + 6 × 11 + 9 × 14 + …
7. 12 + (12 + 22) + (12 + 22 + 32) + …

Ans

$\begin{array}{l}1.\text{}\mathrm{Let}{\text{\hspace{0.17em} S}}_{\text{n}}=\text{1}×\text{2}+\text{2}×\text{3}+\text{3}×\text{4}+\text{4}×\text{5}+...+\mathrm{n}×\left(\mathrm{n}+1\right)\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}{\mathrm{a}}_{\mathrm{n}}=\mathrm{n}×\left(\mathrm{n}+1\right)\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}{\mathrm{S}}_{\mathrm{n}}=\sum \mathrm{n}×\left(\mathrm{n}+1\right)\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\sum {\mathrm{n}}^{2}+\sum \mathrm{n}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\frac{\mathrm{n}\left(\mathrm{n}+1\right)\left(2\mathrm{n}+1\right)}{6}+\frac{\mathrm{n}\left(\mathrm{n}+1\right)}{2}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\frac{\mathrm{n}\left(\mathrm{n}+1\right)}{2}\left(\frac{2\mathrm{n}+1}{3}+1\right)\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\frac{\mathrm{n}\left(\mathrm{n}+1\right)}{2}\left(\frac{2\mathrm{n}+4}{3}\right)\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\frac{\mathrm{n}\left(\mathrm{n}+1\right)}{2}×2\left(\frac{\mathrm{n}+2}{3}\right)\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}{\mathrm{S}}_{\mathrm{n}}\text{\hspace{0.17em}}=\frac{1}{3}\mathrm{n}\left(\mathrm{n}+1\right)\left(\mathrm{n}+2\right)\\ \text{2.}\mathrm{Let}{\text{\hspace{0.17em} S}}_{\text{n}}=\text{1}×\text{2}×\text{3}+\text{2}×\text{3}×\text{4}+\text{3}×\text{4}×\text{5}+...+\mathrm{n}×\left(\mathrm{n}+1\right)×\left(\mathrm{n}+2\right)\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}{\mathrm{a}}_{\mathrm{n}}=\mathrm{n}×\left(\mathrm{n}+1\right)×\left(\mathrm{n}+2\right)\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}{\mathrm{S}}_{\mathrm{n}}=\sum \mathrm{n}×\left(\mathrm{n}+1\right)×\left(\mathrm{n}+2\right)\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\sum \left({\mathrm{n}}^{3}+3{\mathrm{n}}^{2}+2\mathrm{n}\right)\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\sum {\mathrm{n}}^{3}+3\sum {\mathrm{n}}^{2}+2\sum \mathrm{n}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}={\left\{\frac{\mathrm{n}\left(\mathrm{n}+1\right)}{2}\right\}}^{2}+3.\frac{\mathrm{n}\left(\mathrm{n}+1\right)\left(2\mathrm{n}+1\right)}{6}+2.\frac{\mathrm{n}\left(\mathrm{n}+1\right)}{2}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\frac{{\mathrm{n}}^{2}{\left(\mathrm{n}+1\right)}^{2}}{4}+\frac{\mathrm{n}\left(\mathrm{n}+1\right)\left(2\mathrm{n}+1\right)}{2}+\frac{2\mathrm{n}\left(\mathrm{n}+1\right)}{2}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\frac{\mathrm{n}\left(\mathrm{n}+1\right)}{2}\left\{\frac{\mathrm{n}\left(\mathrm{n}+1\right)}{2}+\left(2\mathrm{n}+1\right)+2\right\}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\frac{\mathrm{n}\left(\mathrm{n}+1\right)}{2}\left\{\frac{{\mathrm{n}}^{2}+\mathrm{n}+4\mathrm{n}+2+4}{2}\right\}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\frac{\mathrm{n}\left(\mathrm{n}+1\right)}{2}\left(\frac{{\mathrm{n}}^{2}+5\mathrm{n}+6}{2}\right)\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\frac{\mathrm{n}\left(\mathrm{n}+1\right)\left(\mathrm{n}+2\right)\left(\mathrm{n}+3\right)}{4}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}{\mathrm{S}}_{\mathrm{n}}\text{\hspace{0.17em}}=\frac{1}{4}\mathrm{n}\left(\mathrm{n}+1\right)\left(\mathrm{n}+2\right)\left(\mathrm{n}+3\right)\\ 3.{\mathrm{a}}_{\mathrm{n}}=\left(2\text{\hspace{0.17em}}\mathrm{n}+1\right)×{\mathrm{n}}^{2}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}{\mathrm{S}}_{\mathrm{n}}=\sum \left(2{\mathrm{n}}^{3}+{\mathrm{n}}^{2}\right)\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=2\text{\hspace{0.17em}}\sum {\mathrm{n}}^{3}+\sum {\mathrm{n}}^{2}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=2{\left\{\frac{\mathrm{n}\left(\mathrm{n}+1\right)}{2}\right\}}^{2}+\frac{\mathrm{n}\left(\mathrm{n}+1\right)\left(2\mathrm{n}+1\right)}{6}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=2.\frac{{\mathrm{n}}^{2}{\left(\mathrm{n}+1\right)}^{2}}{4}+\frac{\mathrm{n}\left(\mathrm{n}+1\right)\left(2\mathrm{n}+1\right)}{6}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\frac{{\mathrm{n}}^{2}{\left(\mathrm{n}+1\right)}^{2}}{2}+\frac{\mathrm{n}\left(\mathrm{n}+1\right)\left(2\mathrm{n}+1\right)}{6}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\frac{\mathrm{n}\left(\mathrm{n}+1\right)}{2}\left\{\mathrm{n}\left(\mathrm{n}+1\right)+\frac{\left(2\mathrm{n}+1\right)}{3}\right\}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\frac{\mathrm{n}\left(\mathrm{n}+1\right)}{2}\left\{\frac{3\mathrm{n}\left(\mathrm{n}+1\right)+\left(2\mathrm{n}+1\right)}{3}\right\}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\frac{\mathrm{n}\left(\mathrm{n}+1\right)}{2}\left\{\frac{3{\mathrm{n}}^{2}+3\mathrm{n}+2\mathrm{n}+1}{3}\right\}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\frac{\mathrm{n}\left(\mathrm{n}+1\right)}{2}\left\{\frac{3{\mathrm{n}}^{2}+5\mathrm{n}+1}{3}\right\}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}{\mathrm{S}}_{\mathrm{n}}\text{\hspace{0.17em}}=\frac{1}{6}\mathrm{n}\left(\mathrm{n}+1\right)\left(3{\mathrm{n}}^{2}+5\mathrm{n}+1\right)\\ 4.\text{}\mathrm{Let}\text{\hspace{0.17em}\hspace{0.17em}}{\mathrm{S}}_{\mathrm{n}}=\frac{1}{1×2}+\frac{1}{2×3}+\frac{1}{3×4}+...+\frac{1}{\mathrm{n}\left(\mathrm{n}+1\right)}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\left(\frac{1}{1}-\frac{1}{2}\right)+\left(\frac{1}{2}-\frac{1}{3}\right)+\left(\frac{1}{3}-\frac{1}{4}\right)+...+\left(\frac{1}{\mathrm{n}}-\frac{1}{\mathrm{n}+1}\right)\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\text{\hspace{0.17em}}\frac{1}{1}-\frac{1}{\mathrm{n}+1}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\frac{\mathrm{n}+1-1}{\mathrm{n}+1}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\frac{\mathrm{n}}{\mathrm{n}+1}\\ \text{5.}{5}^{2}+{6}^{2}+{7}^{2}+...\text{}+\text{}{20}^{2}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}={1}^{2}+{2}^{2}+{3}^{2}+{4}^{2}+{5}^{2}+{6}^{2}+{7}^{2}+...\text{}+\text{}{20}^{2}-\left({1}^{2}+{2}^{2}+{3}^{2}+{4}^{2}\right)\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\sum {20}^{2}-\sum {4}^{2}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\frac{20\left(20+1\right)\left(2×20+1\right)}{6}-\frac{4\left(4+1\right)\left(2×4+1\right)}{6}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\frac{20×21×41}{6}-\frac{4×5×9}{6}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\frac{17220-180}{6}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\frac{17040}{6}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=2840\\ 6.{\text{Let\hspace{0.17em}S}}_{\text{n}}=\text{3}×\text{8}+\text{6}×\text{11}+\text{9}×\text{14}+...\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}{\mathrm{T}}_{\mathrm{n}}=\left(3,6,9,...{\mathrm{n}}^{\mathrm{th}}\text{\hspace{0.17em}}\mathrm{term}\right)\left(8,11,14,...{\text{n}}^{\text{th}}\text{term}\right)\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\left\{3+\left(\mathrm{n}-1\right)3\right\}\left\{8+\left(\mathrm{n}-1\right)3\right\}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\left(3\text{\hspace{0.17em}}\mathrm{n}\right)\left(3\text{\hspace{0.17em}}\mathrm{n}+5\right)\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=9\text{\hspace{0.17em}}{\mathrm{n}}^{2}+15\mathrm{n}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}{\mathrm{S}}_{\mathrm{n}}=\sum {\mathrm{T}}_{\mathrm{n}}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\sum \left(9{\mathrm{n}}^{2}+15\mathrm{n}\right)\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=9\sum {\mathrm{n}}^{2}+15\sum \mathrm{n}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=9×\frac{\mathrm{n}\left(\mathrm{n}+1\right)\left(2\mathrm{n}+1\right)}{6}+15×\frac{\mathrm{n}\left(\mathrm{n}+1\right)}{2}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=3×\frac{\mathrm{n}\left(\mathrm{n}+1\right)\left(2\mathrm{n}+1\right)}{2}+15×\frac{\mathrm{n}\left(\mathrm{n}+1\right)}{2}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\frac{3\mathrm{n}\left(\mathrm{n}+1\right)}{2}\left\{\left(2\mathrm{n}+1\right)+5\right\}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\frac{3\mathrm{n}\left(\mathrm{n}+1\right)}{2}\left(2\mathrm{n}+6\right)\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=3\mathrm{n}\left(\mathrm{n}+1\right)\left(\mathrm{n}+3\right)\\ 7.\text{}\mathrm{Let}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}{\mathrm{S}}_{\mathrm{n}}={\text{1}}^{\text{2}}+\left({\text{1}}^{\text{2}}+{\text{2}}^{\text{2}}\right)+\left({\text{1}}^{\text{2}}+{\text{2}}^{\text{2}}+{\text{3}}^{\text{2}}\right)+\text{}...\text{\hspace{0.17em}\hspace{0.17em}}+\left({\text{1}}^{\text{2}}+{\text{2}}^{\text{2}}+{\text{3}}^{\text{2}}+...+{\mathrm{n}}^{2}\right)\\ \mathrm{Here},{\text{\hspace{0.17em}T}}_{\text{n}}={\text{1}}^{\text{2}}+{\text{2}}^{\text{2}}+{\text{3}}^{\text{2}}+...+{\mathrm{n}}^{2}\\ =\sum {\mathrm{n}}^{2}\\ =\frac{\mathrm{n}\left(\mathrm{n}+1\right)\left(2\mathrm{n}+1\right)}{6}\\ =\frac{1}{6}\mathrm{n}\left(2{\mathrm{n}}^{2}+3\mathrm{n}+1\right)\\ =\frac{1}{6}\left(2{\mathrm{n}}^{3}+3{\mathrm{n}}^{2}+\mathrm{n}\right)\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}{\mathrm{S}}_{\mathrm{n}}=\sum {\mathrm{T}}_{\mathrm{n}}\\ =\frac{1}{6}\sum \left(2{\mathrm{n}}^{3}+3{\mathrm{n}}^{2}+\mathrm{n}\right)\\ =\frac{1}{6}\left(2\sum {\mathrm{n}}^{3}+3\sum {\mathrm{n}}^{2}+\sum \mathrm{n}\right)\\ =\frac{1}{6}\left[2{\left\{\frac{\mathrm{n}\left(\mathrm{n}+1\right)}{2}\right\}}^{2}+3\frac{\mathrm{n}\left(\mathrm{n}+1\right)\left(2\mathrm{n}+1\right)}{6}+\frac{\mathrm{n}\left(\mathrm{n}+1\right)}{2}\right]\\ =\frac{1}{6}\left\{\frac{{\mathrm{n}}^{2}{\left(\mathrm{n}+1\right)}^{2}}{2}+\frac{\mathrm{n}\left(\mathrm{n}+1\right)\left(2\mathrm{n}+1\right)}{2}+\frac{\mathrm{n}\left(\mathrm{n}+1\right)}{2}\right\}\\ =\frac{1}{6}×\frac{\mathrm{n}\left(\mathrm{n}+1\right)}{2}\left\{\mathrm{n}\left(\mathrm{n}+1\right)+\left(2\mathrm{n}+1\right)+1\right\}\\ =\frac{1}{6}×\frac{\mathrm{n}\left(\mathrm{n}+1\right)}{2}\left({\mathrm{n}}^{2}+\mathrm{n}+2\mathrm{n}+2\right)\\ =\frac{1}{12}\mathrm{n}\left(\mathrm{n}+1\right)\left({\mathrm{n}}^{2}+3\mathrm{n}+2\right)\\ =\frac{1}{12}\mathrm{n}\left(\mathrm{n}+1\right)\left(\mathrm{n}+1\right)\left(\mathrm{n}+2\right)\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}{\mathrm{S}}_{\mathrm{n}}=\frac{\mathrm{n}{\left(\mathrm{n}+1\right)}^{2}\left(\mathrm{n}+2\right)}{12}\end{array}$

Q.57 Find the sum to n terms of the series in whose nth terms is given by:

1. n (n + 1) (n + 4).
2. n2 + 2n
3. (2n – 1)2

Ans

1.

${\text{Let\hspace{0.17em}\hspace{0.17em}T}}_{\text{n}}=\text{n}\left(\text{n}+\text{1}\right)\left(\text{n}+\text{4}\right)$ $\begin{array}{l}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\mathrm{n}\left({\mathrm{n}}^{2}+5\mathrm{n}+4\right)\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}={\mathrm{n}}^{3}+5{\mathrm{n}}^{2}+4\mathrm{n}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}{\mathrm{S}}_{\mathrm{n}}=\sum {\mathrm{T}}_{\mathrm{n}}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\sum \left({\mathrm{n}}^{3}+5{\mathrm{n}}^{2}+4\mathrm{n}\right)\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\sum {\mathrm{n}}^{3}+5\sum {\mathrm{n}}^{2}+4\sum \mathrm{n}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}={\left\{\frac{\mathrm{n}\left(\mathrm{n}+1\right)}{2}\right\}}^{2}+\frac{5\text{\hspace{0.17em}}\mathrm{n}\left(\mathrm{n}+1\right)\left(2\mathrm{n}+1\right)}{6}+\frac{4\mathrm{n}\left(\mathrm{n}+1\right)}{2}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\frac{\mathrm{n}\left(\mathrm{n}+1\right)}{2}\left\{\frac{\mathrm{n}\left(\mathrm{n}+1\right)}{2}+5.\frac{\left(2\mathrm{n}+1\right)}{3}+4\right\}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\frac{\mathrm{n}\left(\mathrm{n}+1\right)}{2}\left\{\frac{3\mathrm{n}\left(\mathrm{n}+1\right)+10\left(2\mathrm{n}+1\right)+24}{6}\right\}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\frac{\mathrm{n}\left(\mathrm{n}+1\right)}{2}\left(\frac{3{\mathrm{n}}^{2}+3\mathrm{n}+20\text{\hspace{0.17em}}\mathrm{n}+10+24}{6}\right)\\ {\mathrm{S}}_{\mathrm{n}}=\frac{\mathrm{n}\left(\mathrm{n}+1\right)}{12}\left(3{\mathrm{n}}^{2}+23\text{\hspace{0.17em}}\mathrm{n}+34\right)\end{array}$

2.

$\begin{array}{l}\mathrm{Let}{\text{\hspace{0.17em}\hspace{0.17em}T}}_{\text{n}}={\text{n}}^{\text{2}}+{\text{2}}^{\text{n}}\\ {\mathrm{S}}_{\mathrm{n}}=\sum {\mathrm{T}}_{\mathrm{n}}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\sum \left({\text{n}}^{\text{2}}+{\text{2}}^{\text{n}}\right)\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\sum {\text{n}}^{\text{2}}+\sum {\text{2}}^{\text{n}}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\frac{\mathrm{n}\left(\mathrm{n}+1\right)\left(2\mathrm{n}+1\right)}{6}+\left(2+{2}^{2}+{2}^{3}+{2}^{4}+...+{2}^{\mathrm{n}}\right)\end{array}$ $\begin{array}{l}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\end{array}$ $\begin{array}{l}=\frac{\mathrm{n}\left(\mathrm{n}+1\right)\left(2\mathrm{n}+1\right)}{6}+\frac{2\left({2}^{\mathrm{n}}-1\right)}{2-1}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\left[âˆµ{\mathrm{S}}_{\mathrm{n}}=\frac{\mathrm{a}\left({\mathrm{r}}^{\mathrm{n}}-1\right)}{\mathrm{r}-1}\right]\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\frac{\mathrm{n}\left(\mathrm{n}+1\right)\left(2\mathrm{n}+1\right)}{6}+2\left({2}^{\mathrm{n}}-1\right)\\ \therefore {\mathrm{S}}_{\mathrm{n}}=\frac{1}{6}\mathrm{n}\left(\mathrm{n}+1\right)\left(2\mathrm{n}+1\right)+2\left({2}^{\mathrm{n}}-1\right)\end{array}$

3.

$\begin{array}{l}\mathrm{Let}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}{\mathrm{t}}_{\mathrm{n}}={\left(\text{2n}-\text{1}\right)}^{\text{2}}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=4{\mathrm{n}}^{2}-4\mathrm{n}+1\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}{\mathrm{S}}_{\mathrm{n}}=\sum \left(4{\mathrm{n}}^{2}-4\mathrm{n}+1\right)\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=4\sum {\mathrm{n}}^{2}-4\sum \mathrm{n}+\sum 1\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=4\frac{\mathrm{n}\left(\mathrm{n}+1\right)\left(2\mathrm{n}+1\right)}{6}-4\frac{\mathrm{n}\left(\mathrm{n}+1\right)}{2}+\mathrm{n}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=2\frac{\mathrm{n}\left(\mathrm{n}+1\right)\left(2\mathrm{n}+1\right)}{3}-2\mathrm{n}\left(\mathrm{n}+1\right)+\mathrm{n}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\frac{2\mathrm{n}\left(2{\mathrm{n}}^{2}+3\text{\hspace{0.17em}}\mathrm{n}+1\right)}{3}-2{\mathrm{n}}^{2}-2\mathrm{n}+\mathrm{n}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\frac{2\left(2{\mathrm{n}}^{3}+3\text{\hspace{0.17em}}{\mathrm{n}}^{2}+\mathrm{n}\right)}{3}-2{\mathrm{n}}^{2}-\mathrm{n}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\frac{4{\mathrm{n}}^{3}+6\text{\hspace{0.17em}}{\mathrm{n}}^{2}+2\mathrm{n}-6{\mathrm{n}}^{2}-3\mathrm{n}}{3}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\frac{4{\mathrm{n}}^{3}-\mathrm{n}}{3}\end{array}$ $\begin{array}{l}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\frac{\mathrm{n}}{3}\left(4{\mathrm{n}}^{2}-1\right)\\ {\mathrm{S}}_{\mathrm{n}}=\frac{\mathrm{n}}{3}\left(2\mathrm{n}-1\right)\left(2\mathrm{n}+1\right)\end{array}$

Q.58

$\begin{array}{l}\text{Find the sum to infinity in each of the following Geometric Progression.}\\ \text{1. 1,}\frac{\text{1}}{\text{3}}\text{,}\frac{\text{1}}{\text{9}}\text{,}...\\ \text{2. 6, 1.2, 0.24,}...\\ \text{3. 5,}\frac{\text{20}}{\text{7}}\text{,}\frac{\text{80}}{\text{49}}\text{,}...\\ 4.\text{\hspace{0.17em}\hspace{0.17em}}\frac{-\text{3}}{\text{4}}\text{,}\frac{\text{3}}{\text{16}}\text{,}\frac{\text{-3}}{\text{64}}\text{,}...\\ {\text{5. Prove that:3}}^{\frac{\text{1}}{\text{2}}}{\text{×3}}^{\frac{\text{1}}{\text{4}}}{\text{×3}}^{\frac{\text{1}}{\text{8}}}...\text{=3}\\ {\text{6. Let x = 1+a+a}}^{\text{2}}\text{+}...{\text{and y = 1+b+b}}^{\text{2}}\text{+}...\text{, where}\left|\text{a}\right|\text{<1 and}\left|\text{b}\right|\text{>1.}\\ {\text{Prove that 1+ab+a}}^{\text{2}}{\text{b}}^{\text{2}}\text{+}...\text{=}\frac{\text{xy}}{\text{x+y-1}}\end{array}$

Ans

$\begin{array}{l}1.\text{\hspace{0.17em}Given G.P. is:}\\ \text{1,}\frac{1}{3},\frac{1}{9},...\\ \mathrm{a}=1,\text{r}=\frac{\left(\frac{1}{3}\right)}{1}=\frac{1}{3}\\ {\mathrm{S}}_{\infty }=\frac{\mathrm{a}}{1-\mathrm{r}}\\ =\frac{1}{1-\frac{1}{3}}\\ =\frac{1}{\frac{2}{3}}\\ =\frac{3}{2}=1.5\\ \text{2. The given G.P. is:}\\ \text{6},\text{1}.\text{2},\text{}0.\text{24},\dots \\ \mathrm{a}=6,\text{}\\ \text{r}=\frac{1.2}{6}=0.2\\ {\mathrm{S}}_{\infty }=\frac{\mathrm{a}}{1-\mathrm{r}}\\ =\frac{6}{1-0.2}\\ =\frac{6}{0.8}\\ =\frac{60}{8}\\ =7.5\\ \mathrm{Thus},\text{the sum of infinite terms is 7.5.}\\ \text{3. The given G.P. is:}\\ 5,\frac{20}{7},\frac{80}{49},...\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{a}=5,\text{}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}r}=\frac{\left(\frac{20}{7}\right)}{5}=\frac{4}{7}\\ {\mathrm{S}}_{\infty }=\frac{\mathrm{a}}{1-\mathrm{r}}\\ =\frac{5}{1-\frac{4}{7}}\\ =\frac{5}{\left(\frac{3}{7}\right)}\\ =\frac{35}{3}\\ =11.6\\ \mathrm{Thus},\text{the sum of infinite terms is 11.6.}\\ \text{4. The given G.P. is:}\\ \frac{-3}{4},\frac{3}{16},\frac{-3}{64},...\\ \mathrm{a}=\frac{-3}{4},\text{}\\ \text{\hspace{0.17em}r}=\frac{\left(\frac{3}{16}\right)}{\left(\frac{-3}{4}\right)}\\ =-\frac{1}{4}\\ {\mathrm{S}}_{\infty }=\frac{\mathrm{a}}{1-\mathrm{r}}=\frac{\left(-\frac{3}{4}\right)}{1-\left(\frac{-1}{4}\right)}=-\frac{3}{4}×\frac{4}{5}=-\frac{3}{5}\\ \text{5. L.H.S.=}{3}^{\frac{1}{2}}×{3}^{\frac{1}{4}}×{3}^{\frac{1}{8}}...\\ \text{\hspace{0.17em}}={3}^{\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+...}\\ \text{\hspace{0.17em}}={3}^{\left(\frac{\frac{1}{2}}{1-\frac{1}{2}}\right)}\left[âˆµ\mathrm{S}\infty =\frac{\mathrm{a}}{1-\mathrm{r}}\right]\\ \text{\hspace{0.17em}}={3}^{\left(\frac{\frac{1}{2}}{\frac{1}{2}}\right)}\\ ={3}^{1}\\ =3=\mathrm{R}.\mathrm{H}.\mathrm{S}.\\ \mathrm{Hence}\text{, it was to be proved.}\\ \text{6. x =}1+\mathrm{a}+{\mathrm{a}}^{2}+...\\ \text{\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}}=\frac{1}{1-\mathrm{a}}\left[{\mathrm{S}}_{\infty }=\frac{\mathrm{a}}{1-\mathrm{r}}\right]\\ \text{and y}=1+\mathrm{b}+{\mathrm{b}}^{2}+...\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\frac{1}{1-\mathrm{b}}\\ \mathrm{L}.\mathrm{H}.\mathrm{S}.=1+\mathrm{ab}+{\mathrm{a}}^{2}{\mathrm{b}}^{2}+...\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}}=\frac{1}{1-\mathrm{ab}}\\ \mathrm{R}.\mathrm{H}.\mathrm{S}.=\frac{\mathrm{xy}}{\mathrm{x}+\mathrm{y}-1}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}}=\frac{\left(\frac{1}{1-\mathrm{a}}\right)\left(\frac{1}{1-\mathrm{b}}\right)}{\left(\frac{1}{1-\mathrm{a}}\right)+\left(\frac{1}{1-\mathrm{b}}\right)-1}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}}=\frac{1}{1-\mathrm{b}+1-\mathrm{a}-\left(1-\mathrm{a}\right)\left(1-\mathrm{b}\right)}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\frac{1}{2-\mathrm{b}-\mathrm{a}-\left(1-\mathrm{b}-\mathrm{a}+\mathrm{ab}\right)}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\frac{1}{2-\mathrm{b}-\mathrm{a}-1+\mathrm{b}+\mathrm{a}-\mathrm{ab}}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\frac{1}{1-\mathrm{ab}}\\ \text{So,L.H.S.=R.H.S.}\end{array}$

Q.59 Show that the sum of (m + n)th and (m – n)th terms of an A.P. is equal to twice the mth term.

Ans

$\begin{array}{l}\mathrm{Let}\text{first term of A.P. be a and common difference be d.}\\ \text{Then,}\\ {\text{T}}_{\text{m+n}}+{\text{T}}_{\text{m-n}}=\left\{\mathrm{a}+\left(\mathrm{m}+\mathrm{n}-1\right)\mathrm{d}\right\}+\left\{\mathrm{a}+\left(\mathrm{m}-\mathrm{n}-1\right)\mathrm{d}\right\}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=2\mathrm{a}+\left(\mathrm{m}+\mathrm{n}-1+\mathrm{m}-\mathrm{n}-1\right)\mathrm{d}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=2\mathrm{a}+2\left(\mathrm{m}-1\right)\mathrm{d}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=2\left\{\mathrm{a}+\left(\mathrm{m}-1\right)\mathrm{d}\right\}\\ \text{\hspace{0.17em}\hspace{0.17em}}=2{\mathrm{T}}_{\mathrm{m}}\\ \mathrm{Thus},\\ {\text{T}}_{\text{m+n}}+{\text{T}}_{\text{m-n}}=2{\mathrm{T}}_{\mathrm{m}}\\ \mathrm{Hence}\text{proved.}\end{array}$

Q.60 If the sum of three numbers in A.P., is 24 and their product is 440, find the numbers.

Ans

$\begin{array}{l}\mathrm{Let}\text{three numbers in A.P. are a}-\mathrm{d},\text{}\mathrm{a},\text{}\mathrm{a}+\mathrm{d}.\\ \mathrm{Then},\text{according to first condition:}\end{array}$ $\begin{array}{l}\left(\text{a}-\mathrm{d}\right)+\mathrm{a}+\left(\mathrm{a}+\mathrm{d}\right)=24\\ ⇒\text{\hspace{0.17em}\hspace{0.17em}}3\mathrm{a}=24\\ ⇒\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{a}=\frac{24}{3}\\ =8\\ \mathrm{According}\text{to second condition:}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\left(\text{a}-\mathrm{d}\right)\mathrm{a}\left(\mathrm{a}+\mathrm{d}\right)=440\\ ⇒\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\left(\text{8}-\mathrm{d}\right)8\left(8+\mathrm{d}\right)=440\\ ⇒\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\left(\text{8}-\mathrm{d}\right)\left(8+\mathrm{d}\right)=\frac{440}{8}\\ ⇒\text{\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}}64-{\mathrm{d}}^{2}=55\\ ⇒\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}{\mathrm{d}}^{2}=64-55\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=9\\ ⇒\text{\hspace{0.17em} \hspace{0.17em} \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{d}=±3\\ \mathrm{The}\text{three numbers in A.P. are:}\\ \text{8}-3,8,8+3\text{or 8}+3,\text{\hspace{0.17em}}8,\text{\hspace{0.17em}}8-3\\ ⇒5,\text{\hspace{0.17em}\hspace{0.17em}}8,\text{\hspace{0.17em}\hspace{0.17em}}11\text{or 11,\hspace{0.17em}8,\hspace{0.17em}5}\\ \text{Thus, the three numbers are 5, 8, 11.}\end{array}$

Q.61 Let the sum of n, 2n, 3n terms of an A.P. be S1, S2 and S3, respectively, show that S3 = 3(S2 – S1).

Ans

$\begin{array}{l}\begin{array}{l}âˆµ{\text{S}}_{1}={\text{S}}_{\text{n}}\\ \text{}=\frac{\text{n}}{2}\left\{2\text{a}+\left(\text{n}-1\right)\text{d}\right\}\\ {\text{S}}_{2}={\text{S}}_{2\text{n}}\end{array}\\ \begin{array}{l}\text{}=\frac{2\text{n}}{2}\left\{2\text{a}+\left(2\text{n}-1\right)\text{d}\right\}\\ {\text{S}}_{3}={\text{S}}_{3\text{n}}\\ \text{}=\frac{3\text{n}}{2}\left\{2\text{a}+\left(3\text{n}-1\right)\text{d}\right\}\end{array}\\ \begin{array}{l}\text{R}.\text{H}.\text{S}.=\text{3}\left({\text{S}}_{\text{2}}-{\text{S}}_{\text{1}}\right)\\ \text{}=\text{3}\left[\frac{2\text{n}}{2}\left\{2\text{a}+\left(2\text{n}-1\right)\text{d}\right\}-\frac{\text{n}}{2}\left\{2\text{a}+\left(\text{n}-1\right)\text{d}\right\}\right]\end{array}\\ \text{}=\frac{3\text{n}}{2}\left[2\left\{2\text{a}+\left(2\text{n}-1\right)\text{d}\right\}-\left\{2\text{a}+\left(\text{n}-1\right)\text{d}\right\}\right]\\ \text{}=\frac{3\text{n}}{2}\left[4\text{a}-2\text{a}+\left(4\text{n}-2-\text{n}+1\right)\text{d}\right]\\ \text{}=\frac{3\text{n}}{2}\left\{2\text{a}+\left(3\text{n}-1\right)\text{d}\right\}\\ \begin{array}{l}\text{}={\text{S}}_{3}=\text{L}.\text{H}.\text{S}.\\ \mathrm{Thus},{\text{S}}_{3}=\text{3}\left({\text{S}}_{\text{2}}-{\text{S}}_{\text{1}}\right)\text{is proved.}\end{array}\end{array}$

Q.62 Find the sum of all numbers between 200 and 400 which are divisible by 7.

Ans

$\begin{array}{l}\mathrm{The}\text{numbers between 200 to 400 divisible by 7 are:}\\ 203,\text{\hspace{0.17em}}210,\text{\hspace{0.17em}}217,224,\text{\hspace{0.17em}}231,...,\text{\hspace{0.17em}}399.\\ \mathrm{Here},\text{a}=203,\text{d}=210-203=7,\text{\hspace{0.17em}\hspace{0.17em}}\mathrm{l}=399\\ âˆµ\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{l}=\mathrm{a}+\left(\mathrm{n}-1\right)\mathrm{d}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}399=203+\left(\mathrm{n}-1\right)7\\ ⇒\text{\hspace{0.17em}}\frac{196}{7}=\mathrm{n}-1\end{array}$ $\begin{array}{l}⇒28+1=\mathrm{n}\\ ⇒\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{n}=29\\ \text{\hspace{0.17em}\hspace{0.17em}}{\mathrm{S}}_{\mathrm{n}}=\frac{\mathrm{n}}{2}\left(\mathrm{a}+\mathrm{l}\right)\\ ⇒\text{\hspace{0.17em}}{\mathrm{S}}_{29}=\frac{29}{2}\left(203+399\right)\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\frac{29}{2}×602\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}}=8729\\ \mathrm{Thus},\text{the sum of multiples of 7 between 200}\\ \text{and 400 is 8729.}\end{array}$

Q.63 Find the sum of integers from 1 to 100 that are divisible by 2 or 5.

Ans

$\begin{array}{l}\mathrm{The}\text{numbers from 1 to 100 divisible by 2:}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}2,4,6,8,...,100\\ \mathrm{Here},\text{a}=2,\text{d}=4-2=2,\text{\hspace{0.17em}\hspace{0.17em}}\mathrm{l}=100\\ âˆµ\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{l}=\mathrm{a}+\left(\mathrm{n}-1\right)\mathrm{d}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}100=2+\left(\mathrm{n}-1\right)2\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\frac{98}{2}=\mathrm{n}-1\\ ⇒\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{n}=49+1\\ =50\\ âˆµ\text{\hspace{0.17em}\hspace{0.17em}}{\mathrm{S}}_{\mathrm{n}}=\frac{\mathrm{n}}{2}\left(\mathrm{a}+\mathrm{l}\right)\end{array}$ $\begin{array}{l}\therefore {\mathrm{S}}_{50}=\frac{50}{2}\left(2+100\right)\\ =25×102\\ =2550\\ \mathrm{The}\text{numbers from 1 to 100 divisible by 5:}\\ 5,10,15,20,...,100\\ \mathrm{Here},\text{a}=5,\text{d}=10-5=5,\text{\hspace{0.17em}\hspace{0.17em}}\mathrm{l}=100\\ âˆµ\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{l}=\mathrm{a}+\left(\mathrm{n}-1\right)\mathrm{d}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}100=5+\left(\mathrm{n}-1\right)5\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\frac{95}{5}=\mathrm{n}-1\\ ⇒\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{n}=19+1\\ =20\\ âˆµ\text{\hspace{0.17em}\hspace{0.17em}}{\mathrm{S}}_{\mathrm{n}}=\frac{\mathrm{n}}{2}\left(\mathrm{a}+\mathrm{l}\right)\\ \therefore {\mathrm{S}}_{20}=\frac{20}{2}\left(5+100\right)\\ =10×105\\ =1050\\ \mathrm{The}\text{numbers from 1 to 100 divisible by 2 and 5:}\\ \text{10,20,30,}...\text{,100}\\ \mathrm{Here},\text{a}=10,\text{d}=20-10=10,\text{\hspace{0.17em}\hspace{0.17em}}\mathrm{l}=100\\ âˆµ\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{l}=\mathrm{a}+\left(\mathrm{n}-1\right)\mathrm{d}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}100=10+\left(\mathrm{n}-1\right)10\\ ⇒\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\frac{90}{10}=\mathrm{n}-1\end{array}$ $\begin{array}{l}\end{array}$ $\begin{array}{l}⇒\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{n}=10\\ âˆµ\text{\hspace{0.17em}\hspace{0.17em}}{\mathrm{S}}_{\mathrm{n}}=\frac{\mathrm{n}}{2}\left(\mathrm{a}+\mathrm{l}\right)\\ \therefore {\mathrm{S}}_{10}=\frac{10}{2}\left(10+100\right)\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=5×110\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=550\\ \mathrm{Sum}\text{of numbers divisible by 2 or 5}\\ =\mathrm{Sum}\text{of multiples of 2}+\mathrm{Sum}\text{of multiples of 5}\\ -\mathrm{Sum}\text{of multiples of 10}\\ =2550+1050-550\\ =3050\end{array}$

Q.64 Find the sum of all two digit numbers which when divided by 4, yields 1 as remainder.

Ans

$\begin{array}{l}\mathrm{Two}\text{digit numbers which when divided by 4, yields}\\ \text{1 as remainder are:}\\ 13,\text{\hspace{0.17em}}17,\text{\hspace{0.17em}}21,...,\text{\hspace{0.17em}}97\\ \mathrm{a}=13,\text{\hspace{0.17em}}\mathrm{d}=17-13=4,\text{\hspace{0.17em}}\mathrm{l}=97\\ âˆµ\text{\hspace{0.17em}}\mathrm{l}=\mathrm{a}+\left(\mathrm{n}-1\right)\mathrm{d}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}97=13+\left(\mathrm{n}-1\right)\mathrm{d}\\ 97-13=\left(\mathrm{n}-1\right)4\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\frac{84}{4}=\mathrm{n}-1\\ \mathrm{n}=21+1\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}}=22\end{array}$ $\begin{array}{l}âˆµ\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}{\mathrm{S}}_{\mathrm{n}}=\frac{\mathrm{n}}{2}\left(\mathrm{a}+\mathrm{l}\right)\\ \therefore \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}{\mathrm{S}}_{22}=\frac{22}{2}\left(13+97\right)\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\frac{22}{2}×110\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=1210\\ \mathrm{Thus},\text{the sum of 22 terms is 1210.}\end{array}$

Q.65

$\begin{array}{l}\text{If f is a function satisfying f}\left(\text{x+y}\right)\text{=f}\left(\text{x}\right)\text{f}\left(\text{y}\right)\text{for all x, y}\in \text{N such that f}\left(\text{1}\right)\text{=3 and}\\ \text{}\sum _{\text{x=1}}^{\text{n}}\text{f}\left(\text{x}\right)\text{=120, find the value of n.}\end{array}$

Ans

$\begin{array}{l}\mathrm{Since},\text{\hspace{0.17em}}\mathrm{f}\left(\mathrm{x}+\mathrm{y}\right)=\mathrm{f}\left(\mathrm{x}\right)\mathrm{f}\left(\mathrm{y}\right)\text{, f}\left(1\right)=3\text{\hspace{0.17em}and}\sum _{\mathrm{x}=1}^{\mathrm{n}}\mathrm{f}\left(\mathrm{x}\right)=120.\\ \mathrm{So},\text{\hspace{0.17em}\hspace{0.17em}}\mathrm{f}\left(2\right)=\mathrm{f}\left(1+1\right)\\ =\mathrm{f}\left(1\right)\mathrm{f}\left(1\right)\\ =3×3\\ ={3}^{2}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{f}\left(3\right)=\mathrm{f}\left(2+1\right)\\ =\mathrm{f}\left(2\right)\mathrm{f}\left(1\right)\\ ={3}^{2}×3\\ ={3}^{3}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{f}\left(4\right)=\mathrm{f}\left(2+2\right)\\ =\mathrm{f}\left(2\right)\mathrm{f}\left(2\right)\end{array}$ $\begin{array}{l}={3}^{2}×{3}^{2}\\ ={3}^{4}\\ âˆµ\sum _{\mathrm{x}=1}^{\mathrm{n}}\mathrm{f}\left(\mathrm{x}\right)=120\\ ⇒\mathrm{f}\left(1\right)+\mathrm{f}\left(2\right)+\mathrm{f}\left(3\right)+\mathrm{f}\left(4\right)+...+\mathrm{f}\left(\mathrm{n}\right)=120\\ 3+{3}^{2}+{3}^{3}+{3}^{4}+...+{3}^{\mathrm{n}}=120\end{array}$ $\begin{array}{l}\end{array}$ $\begin{array}{l}⇒\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\frac{3\left({3}^{\mathrm{n}}-1\right)}{3-1}=120\\ ⇒\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}}{3}^{\mathrm{n}}-1=120×\frac{2}{3}\\ ⇒\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}{3}^{\mathrm{n}}=80+1\\ ⇒\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}{3}^{\mathrm{n}}={3}^{4}\\ ⇒\text{\hspace{0.17em}}\mathrm{n}=4\\ \mathrm{Thus},\text{the value of n is 4.}\end{array}$

Q.66 The sum of some terms of G.P. is 315 whose first term and the common ratio are 5 and 2, respectively. Find the last term and the number of terms.

Ans

$\begin{array}{l}\mathrm{First}\text{term of G.P.}=\text{5}\\ \mathrm{Common}\text{difference of G.P.}=\text{2}\\ \text{Let number of terms in G.P.}=\text{n}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}Sum of n terms in G.P.}=315\\ 5×\frac{\left({2}^{\mathrm{n}}-1\right)}{2-1}=315\end{array}$ $\begin{array}{l}⇒{2}^{\mathrm{n}}-1=\frac{315}{5}\\ {2}^{\mathrm{n}}=63+1\\ ⇒{2}^{\mathrm{n}}={2}^{6}\\ ⇒\mathrm{n}=6\\ âˆµ\mathrm{l}={\mathrm{ar}}^{\mathrm{n}-1}\\ \text{\hspace{0.17em}\hspace{0.17em}}=5×{2}^{\left(6-1\right)}\\ \text{\hspace{0.17em} \hspace{0.17em}}=5×32\\ \text{\hspace{0.17em}\hspace{0.17em}}=160\\ \mathrm{Thus},\text{the last term of G.P. is 160 and number of}\\ \text{terms in G.P. is 6.}\end{array}$

Q.67 The first term of a G.P. is 1. The sum of the third term and fifth term is 90. Find the common ratio of G.P.

Ans

$\begin{array}{l}\text{\hspace{0.17em}}\mathrm{First}\text{term of G.P.}=1\\ \mathrm{Let}\text{common ratio}=\mathrm{r}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}}{\mathrm{T}}_{3}+{\mathrm{T}}_{5}=90\\ ⇒{\mathrm{ar}}^{2}+{\mathrm{ar}}^{4}=90\\ ⇒\text{\hspace{0.17em}\hspace{0.17em}}1.{\mathrm{r}}^{2}+1.{\mathrm{r}}^{4}=90\\ ⇒\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}{\mathrm{r}}^{4}+{\mathrm{r}}^{2}=90\\ ⇒\text{\hspace{0.17em}}{\mathrm{r}}^{4}+{\mathrm{r}}^{2}-90=0\\ ⇒\text{\hspace{0.17em}\hspace{0.17em}}\left({\mathrm{r}}^{2}+10\right)\left({\mathrm{r}}^{2}-9\right)=0\\ ⇒\text{\hspace{0.17em}}{\mathrm{r}}^{2}=9,-10\end{array}$ $\begin{array}{l}⇒\mathrm{r}=±3\left[\mathrm{Neglecting}{\text{r}}^{\text{2}}=-10\right]\\ \mathrm{Thus},\text{the common ratio of G.P. is}±3.\end{array}$

Q.68 The sum of three numbers in G.P. is 56. If we subtract 1, 7, 21 from these numbers in that order, we obtain an arithmetic progression. Find the numbers.

Ans

$\begin{array}{l}\mathrm{Let}\text{three terms in G.P. are}\frac{\mathrm{a}}{\mathrm{r}},\text{\hspace{0.17em}}\mathrm{a},\text{\hspace{0.17em}}\mathrm{ar}.\\ \mathrm{Then},\text{}\mathrm{according}\text{}\mathrm{to}\text{first condition, we have}\\ \frac{\mathrm{a}}{\mathrm{r}}+\text{\hspace{0.17em}}\mathrm{a}+\text{\hspace{0.17em}}\mathrm{ar}=56...\left(\mathrm{i}\right)\\ \text{On subtracting 1},\text{7},\text{21 frome each number respectively,}\\ \left(\frac{\text{a}}{\mathrm{r}}-\text{1}\right),\left(\mathrm{a}-7\right),\left(\mathrm{ar}-21\right)\text{are in A.P., then}\\ 2\left(\mathrm{a}-7\right)=\left(\frac{\text{a}}{\mathrm{r}}-\text{1}\right)+\left(\mathrm{ar}-21\right)\\ 2\mathrm{a}-14=\frac{\text{a}}{\mathrm{r}}+\mathrm{ar}-22\\ ⇒\text{\hspace{0.17em}}\frac{\text{a}}{\mathrm{r}}+\mathrm{ar}-2\mathrm{a}=8...\left(\mathrm{ii}\right)\\ \mathrm{From}\text{equation}\left(\mathrm{i}\right)\text{and equation}\left(\mathrm{ii}\right),\text{we get}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}56}-\mathrm{a}-2\mathrm{a}=8\end{array}$ $\begin{array}{l}⇒56-8=3\mathrm{a}\\ ⇒\mathrm{a}=\frac{48}{3}=16\\ \mathrm{Putting}\text{a}=\text{14 in equation}\left(\mathrm{i}\right),\text{\hspace{0.17em}}\mathrm{we}\text{get}\\ \frac{16}{\mathrm{r}}+\text{\hspace{0.17em}}16+\text{\hspace{0.17em}}16\mathrm{r}=56\\ ⇒\frac{1}{\mathrm{r}}+\mathrm{r}=\frac{56-16}{16}\\ ⇒\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\frac{1}{\mathrm{r}}+\mathrm{r}=\frac{40}{16}\\ ⇒\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}2+2{\mathrm{r}}^{2}=5\mathrm{r}\\ ⇒\text{\hspace{0.17em}\hspace{0.17em}}2{\mathrm{r}}^{2}-5\mathrm{r}+2=0\\ ⇒\text{\hspace{0.17em}}2{\mathrm{r}}^{2}-4\mathrm{r}-\mathrm{r}+2=0\\ ⇒2\mathrm{r}\left(\mathrm{r}-2\right)-1\left(\mathrm{r}-2\right)=0\\ ⇒\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}}\left(\mathrm{r}-2\right)\left(2\mathrm{r}-1\right)=0\\ ⇒\text{\hspace{0.17em}}\mathrm{r}=2,\frac{1}{2}\\ \mathrm{So},\text{three numbers when a}=16\text{and r}=2,\frac{1}{2},\\ 16×2,16,\frac{16}{2}\text{or}\frac{\text{16}}{2}\text{,16,16}×\text{2}\\ \mathrm{i}.\mathrm{e}.,\text{32,16,8 or 8,16,32}\\ \text{Thus, the required three numbers are 8,16,32.}\end{array}$

Q.69 A G.P. consists of an even number of terms. If the sum of all the terms is 5 times the sum of terms occupying odd places, then find its common ratio.

Ans

$\begin{array}{l}\mathrm{Let}\text{G.P. containing even number of terms is:}\\ {\text{a, ar, ar}}^{\text{2}}{\text{, ar}}^{\text{3}}\text{,}...{\text{, ar}}^{\text{n}},{\text{ar}}^{\text{n+1}},.\dots ,{\mathrm{ar}}^{2\mathrm{n}-1},{\mathrm{ar}}^{2\mathrm{n}}\\ \mathrm{According}\text{to given condition,}\\ {\mathrm{S}}_{2\mathrm{n}}=5×\left(\mathrm{a}+{\mathrm{ar}}^{2}+{\mathrm{ar}}^{4}+...+{\mathrm{ar}}^{2\mathrm{n}-1}\right)\\ \left(\begin{array}{l}\text{a}+\text{ar}+{\text{ar}}^{\text{2}}+{\text{ar}}^{\text{3}}+...+{\text{ar}}^{\text{n}}\\ +{\text{ar}}^{\text{n+1}}+...+{\mathrm{ar}}^{2\mathrm{n}-1}+{\mathrm{ar}}^{2\mathrm{n}}\end{array}\right)=5×\left(\mathrm{a}+{\mathrm{ar}}^{2}+{\mathrm{ar}}^{4}+...+{\mathrm{ar}}^{2\mathrm{n}-1}\right)\end{array}$ $\begin{array}{l}\left(\text{ar}+{\text{ar}}^{\text{3}}+...+{\text{ar}}^{\text{n}}+...+{\mathrm{ar}}^{2\mathrm{n}}\right)=4×\left(\mathrm{a}+{\mathrm{ar}}^{2}+{\mathrm{ar}}^{4}+...+{\mathrm{ar}}^{2\mathrm{n}-1}\right)\\ \frac{\mathrm{ar}\left({\mathrm{r}}^{2\mathrm{n}}-1\right)}{\mathrm{r}-1}=4×\frac{\mathrm{a}\left\{{\left({\mathrm{r}}^{2}\right)}^{\mathrm{n}}-1\right\}}{\mathrm{r}-1}\\ ⇒\mathrm{r}\left({\mathrm{r}}^{2\mathrm{n}}-1\right)=4\left({\mathrm{r}}^{2\mathrm{n}}-1\right)\\ ⇒\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{r}=4\\ \mathrm{Thus},\text{the common ratio of G.P. is 4.}\end{array}$

Q.70 The sum of the first four terms of an A.P. is 56. The sum of the last four terms is 112. If its first term is 11, then find the number of terms.

Ans

$\begin{array}{l}\mathrm{Let}\text{common difference of an A.P.}=\text{d}\\ \text{Let number of terms in an A}.\text{P}.=\mathrm{n}\\ \mathrm{The}\text{first term of A.P.}=\text{11}\\ \text{Sum of the first four terms of an A}.\text{P}.=56\end{array}$ $\begin{array}{l}\frac{4}{2}\left\{2×11+\left(4-1\right)\mathrm{d}\right\}=56\left[\mathrm{Sn}=\frac{\mathrm{n}}{2}\left\{2\mathrm{a}+\left(\mathrm{n}-1\right)\mathrm{d}\right\}\right]\\ 3\mathrm{d}=28-22\\ \mathrm{d}=\frac{6}{3}\\ =2\\ \text{Sum of the last four terms\hspace{0.17em}of an A}.\text{P}.=112\\ \left\{\mathrm{a}+\left(\mathrm{n}-4\right)\mathrm{d}\right\}+\left\{\mathrm{a}+\left(\mathrm{n}-3\right)\mathrm{d}\right\}+\left\{\mathrm{a}+\left(\mathrm{n}-2\right)\mathrm{d}\right\}\\ +\left\{\mathrm{a}+\left(\mathrm{n}-1\right)\mathrm{d}\right\}=112\\ \left\{11+\left(\mathrm{n}-4\right)2\right\}+\left\{11+\left(\mathrm{n}-3\right)2\right\}+\left\{11+\left(\mathrm{n}-2\right)2\right\}\\ +\left\{11+\left(\mathrm{n}-1\right)2\right\}=112\\ 44+\left(\mathrm{n}-4+\mathrm{n}-3+\mathrm{n}-2+\mathrm{n}-1\right)2=112\\ 4\mathrm{n}-10=\frac{112-44}{2}\\ \mathrm{n}=\frac{34+10}{4}\\ \text{\hspace{0.17em}\hspace{0.17em}}=11\\ \mathrm{Thus},\text{the number of terms in an A.P. is 11.}\end{array}$

Q.71

$\mathbf{If}\mathbf{\text{}}\mathbf{}\frac{\mathbf{a}\mathbf{+}\mathbf{bx}}{\mathbf{a}\mathbf{-}\mathbf{bx}}\mathbf{=}\frac{\mathbf{b}\mathbf{+}\mathbf{cx}}{\mathbf{b}\mathbf{-}\mathbf{cx}}\mathbf{=}\frac{\mathbf{c}\mathbf{+}\mathbf{dx}}{\mathbf{c}\mathbf{-}\mathbf{dx}}\left(x\ne 0\right)\mathbf{,}\mathbf{}\mathbf{\text{then show that a, b, c and d are in G.P}}\mathbf{.}$

Ans

$\begin{array}{l}\mathrm{We}\text{are given:}\\ \frac{\mathrm{a}+\mathrm{bx}}{\mathrm{a}-\mathrm{bx}}=\frac{\mathrm{b}+\mathrm{cx}}{\mathrm{b}-\mathrm{cx}}=\frac{\mathrm{c}+\mathrm{dx}}{\mathrm{c}-\mathrm{dx}}\left(\mathrm{x}\ne 0\right)\end{array}$ $\begin{array}{l}⇒\frac{\mathrm{a}+\mathrm{bx}}{\mathrm{a}-\mathrm{bx}}=\frac{\mathrm{b}+\mathrm{cx}}{\mathrm{b}-\mathrm{cx}}\\ ⇒\left(\mathrm{a}+\mathrm{bx}\right)\left(\mathrm{b}-\mathrm{cx}\right)=\left(\mathrm{b}+\mathrm{cx}\right)\left(\mathrm{a}-\mathrm{bx}\right)\\ ⇒\mathrm{ab}-\mathrm{acx}+{\mathrm{b}}^{2}\mathrm{x}-{\mathrm{bcx}}^{2}=\mathrm{ba}-{\mathrm{b}}^{2}\mathrm{x}+\mathrm{cax}-{\mathrm{cbx}}^{2}\\ ⇒2{\mathrm{b}}^{2}\mathrm{x}=2\mathrm{acx}\\ ⇒\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}{\mathrm{b}}^{2}=\mathrm{ac}\\ ⇒\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}}\frac{\mathrm{b}}{\mathrm{a}}=\frac{\mathrm{c}}{\mathrm{b}}...\left(\mathrm{i}\right)\\ \mathrm{And}\text{\hspace{0.17em}}\frac{\mathrm{b}+\mathrm{cx}}{\mathrm{b}-\mathrm{cx}}=\frac{\mathrm{c}+\mathrm{dx}}{\mathrm{c}-\mathrm{dx}}\\ ⇒\left(\mathrm{b}+\mathrm{cx}\right)\left(\mathrm{c}-\mathrm{dx}\right)=\left(\mathrm{c}+\mathrm{dx}\right)\left(\mathrm{b}-\mathrm{cx}\right)\\ ⇒\mathrm{bc}-\mathrm{bdx}+{\mathrm{c}}^{2}\mathrm{x}-{\mathrm{cdx}}^{2}=\mathrm{cb}-{\mathrm{c}}^{2}\mathrm{x}+\mathrm{dbx}-{\mathrm{dcx}}^{2}\\ ⇒2{\mathrm{c}}^{2}\mathrm{x}=2\mathrm{bdx}\\ ⇒\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}{\mathrm{c}}^{2}=\mathrm{bd}\\ ⇒\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\frac{\mathrm{c}}{\mathrm{b}}=\frac{\mathrm{d}}{\mathrm{c}}...\left(\mathrm{ii}\right)\\ \mathrm{From}\text{equation}\left(\mathrm{i}\right)\text{and equation}\left(\mathrm{ii}\right),\text{we get}\\ \frac{\mathrm{b}}{\mathrm{a}}=\frac{\mathrm{c}}{\mathrm{b}}=\frac{\mathrm{d}}{\mathrm{c}}\\ ⇒\mathrm{a},\mathrm{b},\mathrm{c},\mathrm{d}\text{are in G.P.}\end{array}$

Q.72 Let S be the sum, P the product and R the sum of reciprocals of n terms in a G.P. Prove that P2Rn = Sn.

Ans

$\begin{array}{l}\mathrm{Let}\text{n terms of G.P. are}\\ {\text{a, ar, ar}}^{\text{2}}{\text{, ar}}^{\text{3}}\text{,}...{\text{, ar}}^{\text{n}-3},{\text{ar}}^{\text{n}-2},{\text{ar}}^{\text{n}-1}\end{array}$ $\begin{array}{l}\mathrm{Then}\text{according to the given conditions:}\\ \text{S}=\text{a}+\text{ar}+{\text{ar}}^{\text{2}}+{\text{ar}}^{\text{3}}+...+{\text{ar}}^{\text{n}-3}+{\text{ar}}^{\text{n}-2}+{\text{ar}}^{\text{n}-1}\\ =\frac{\mathrm{a}\left({\mathrm{r}}^{\mathrm{n}}-1\right)}{\left(\mathrm{r}-1\right)}\\ \mathrm{P}=\text{a}×\text{ar}×{\text{ar}}^{\text{2}}×{\text{ar}}^{\text{3}}×...×{\text{ar}}^{\text{n}-3}×{\text{ar}}^{\text{n}-2}×{\text{ar}}^{\text{n}-1}\\ ={\mathrm{a}}^{\mathrm{n}}×{\mathrm{r}}^{1+2+3+...+\left(\mathrm{n}-3\right)+\left(\mathrm{n}-2\right)+\left(\mathrm{n}-1\right)}\\ ={\mathrm{a}}^{\mathrm{n}}×{\mathrm{r}}^{\frac{\left(\mathrm{n}-1\right)\left(1+\mathrm{n}-1\right)}{2}}\\ ={\mathrm{a}}^{\mathrm{n}}×{\mathrm{r}}^{\frac{\mathrm{n}\left(\mathrm{n}-1\right)}{2}}\\ \mathrm{Squarring}\text{both sides, we get}\\ {\text{P}}^{\text{2}}={\mathrm{a}}^{2\mathrm{n}}×{\mathrm{r}}^{\mathrm{n}\left(\mathrm{n}-1\right)}\\ \mathrm{R}=\frac{1}{\mathrm{a}}+\frac{1}{\mathrm{ar}}+\frac{1}{{\mathrm{ar}}^{2}}+\frac{1}{{\mathrm{ar}}^{3}}+...+\frac{1}{{\mathrm{ar}}^{\mathrm{n}-1}}\\ =\frac{\frac{1}{\mathrm{a}}\left(1-\frac{1}{{\mathrm{r}}^{\mathrm{n}}}\right)}{\left(1-\frac{1}{\mathrm{r}}\right)}\\ =\frac{\mathrm{r}}{{\mathrm{ar}}^{\mathrm{n}}}\left(\frac{{\mathrm{r}}^{\mathrm{n}}-1}{\mathrm{r}-1}\right)\\ \therefore {\mathrm{R}}^{\mathrm{n}}=\frac{{\mathrm{r}}^{\mathrm{n}\left(1-\mathrm{n}\right)}}{{\mathrm{a}}^{\mathrm{n}}}{\left(\frac{{\mathrm{r}}^{\mathrm{n}}-1}{\mathrm{r}-1}\right)}^{\mathrm{n}}\end{array}$ $\begin{array}{l}\mathrm{L}.\mathrm{H}.\mathrm{S}.={\mathrm{P}}^{2}{\mathrm{R}}^{\mathrm{n}}\\ ={\mathrm{a}}^{2\mathrm{n}}×{\mathrm{r}}^{\mathrm{n}\left(\mathrm{n}-1\right)}×\frac{{\mathrm{r}}^{\mathrm{n}\left(1-\mathrm{n}\right)}}{{\mathrm{a}}^{\mathrm{n}}}{\left(\frac{{\mathrm{r}}^{\mathrm{n}}-1}{\mathrm{r}-1}\right)}^{\mathrm{n}}\\ ={\mathrm{a}}^{\mathrm{n}}×{{\mathrm{r}}^{{\mathrm{n}}^{2}-\mathrm{n}}}^{+\mathrm{n}-{\mathrm{n}}^{2}}×{\left(\frac{{\mathrm{r}}^{\mathrm{n}}-1}{\mathrm{r}-1}\right)}^{\mathrm{n}}\end{array}$ $\begin{array}{l}\text{}\end{array}$ $\begin{array}{l}={\mathrm{a}}^{\mathrm{n}}×{\left(\frac{{\mathrm{r}}^{\mathrm{n}}-1}{\mathrm{r}-1}\right)}^{\mathrm{n}}\\ ={\left\{\frac{\mathrm{a}\left({\mathrm{r}}^{\mathrm{n}}-1\right)}{\mathrm{r}-1}\right\}}^{\mathrm{n}}\\ ={\mathrm{S}}^{\mathrm{n}}=\mathrm{R}.\mathrm{H}.\mathrm{S}.\end{array}$

Q.73 The pth, qth and rth terms of an A.P. are a, b, c, respectively. Show that (q – r )a + (r – p )b + (p – q )c = 0.

Ans

$\begin{array}{l}\mathrm{Let}\text{first term of A.P. is A and common difference is D.}\\ \text{Then, according to given condition:}\\ \text{\hspace{0.17em} \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}{\mathrm{T}}_{\mathrm{p}}=\mathrm{a}\\ ⇒\mathrm{A}+\left(\mathrm{p}-1\right)\mathrm{D}=\mathrm{a}...\left(\mathrm{i}\right)\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}{\mathrm{T}}_{\mathrm{q}}=\mathrm{b}\\ ⇒\mathrm{A}+\left(\mathrm{q}-1\right)\mathrm{D}=\mathrm{b}\text{\hspace{0.17em}}...\left(\mathrm{ii}\right)\\ \text{\hspace{0.17em} \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}{\mathrm{T}}_{\mathrm{r}}=\mathrm{c}\\ ⇒\mathrm{A}+\left(\mathrm{r}-1\right)\mathrm{D}=\mathrm{c}\text{\hspace{0.17em}}...\left(\mathrm{iii}\right)\end{array}$ $\begin{array}{l}\mathrm{L}.\mathrm{H}.\mathrm{S}.=\left(\text{q}-\text{r}\right)\text{a}+\left(\text{r}-\text{p}\right)\text{b}+\left(\text{p}-\text{q}\right)\text{c}\\ =\left(\text{q}-\text{r}\right)\left\{\mathrm{A}+\left(\mathrm{p}-1\right)\mathrm{D}\right\}+\left(\text{r}-\text{p}\right)\left\{\mathrm{A}+\left(\mathrm{q}-1\right)\mathrm{D}\right\}\\ +\left(\text{p}-\text{q}\right)\left\{\mathrm{A}+\left(\mathrm{r}-1\right)\mathrm{D}\right\}\\ =\mathrm{A}\left(\text{q}-\text{r}+\text{r}-\text{p}+\text{p}-\text{q}\right)+\mathrm{D}\left\{\begin{array}{l}\left(\text{q}-\text{r}\right)\left(\mathrm{p}-1\right)+\left(\text{r}-\text{p}\right)\left(\mathrm{q}-1\right)\\ +\left(\text{p}-\text{q}\right)\left(\mathrm{r}-1\right)\end{array}\right\}\\ =\mathrm{A}\left(0\right)+\mathrm{D}\left(\mathrm{qp}-\mathrm{q}-\mathrm{rp}+\mathrm{r}+\mathrm{rq}-\mathrm{r}-\mathrm{pq}+\mathrm{p}+\mathrm{pr}-\mathrm{p}-\mathrm{qr}+\mathrm{q}\right)\\ =0+0\\ =0=\mathrm{R}.\mathrm{H}.\mathrm{S}.\\ \mathrm{This}\text{was to be proved.}\end{array}$

Q.74

$\mathbf{\text{If a}}\left(\frac{\text{1}}{\text{b}}\text{+}\frac{\text{1}}{\text{c}}\right)\mathbf{\text{,\hspace{0.17em}b}}\left(\frac{\text{1}}{\text{c}}\text{+}\frac{\text{1}}{\text{a}}\right)\mathbf{\text{,\hspace{0.17em}c}}\left(\frac{\text{1}}{\text{a}}\text{+}\frac{\text{1}}{\text{b}}\right)\mathbf{\text{are in A.P., prove that a, b and c are in A.P.}}$

Ans

$\begin{array}{l}\text{Since,a}\left(\frac{\text{1}}{\text{b}}\text{+}\frac{\text{1}}{\text{c}}\right)\text{,\hspace{0.17em}b}\left(\frac{\text{1}}{\text{c}}\text{+}\frac{\text{1}}{\text{a}}\right)\text{,c}\left(\frac{\text{1}}{\text{a}}\text{+}\frac{\text{1}}{\text{b}}\right)\text{areinA.P.}\\ ⇒\text{}\left(\frac{\text{a}}{\text{b}}\text{+}\frac{\text{a}}{\text{c}}\right)\text{,}\left(\frac{\text{b}}{\text{c}}\text{+}\frac{\text{b}}{\text{a}}\right)\text{,\hspace{0.17em}}\left(\frac{\text{c}}{\text{a}}\text{+}\frac{\text{c}}{\text{b}}\right)\text{areinA.P.}\\ ⇒\text{}\left(\frac{\text{a}}{\text{a}}\text{+}\frac{\text{a}}{\text{b}}\text{+}\frac{\text{a}}{\text{c}}\right)\text{,}\left(\frac{\text{b}}{\text{b}}\text{+}\frac{\text{b}}{\text{c}}\text{+}\frac{\text{b}}{\text{a}}\right)\text{,\hspace{0.17em}}\left(\frac{\text{c}}{\text{c}}\text{+}\frac{\text{c}}{\text{a}}\text{+}\frac{\text{c}}{\text{b}}\right)\text{areinA.P.}\\ ⇒\text{a}\left(\frac{\text{1}}{\text{a}}\text{+}\frac{\text{1}}{\text{b}}\text{+}\frac{\text{1}}{\text{c}}\right)\text{,b}\left(\frac{\text{1}}{\text{b}}\text{+}\frac{\text{1}}{\text{c}}\text{+}\frac{\text{1}}{\text{a}}\right)\text{,\hspace{0.17em}c}\left(\frac{\text{1}}{\text{c}}\text{+}\frac{\text{1}}{\text{a}}\text{+}\frac{\text{1}}{\text{b}}\right)\text{areinA.P.}\\ \mathrm{⇒}\text{a,b,careinA.P.}\end{array}$

Q.75 If a, b, c, d are in G.P, prove that (an + bn), (bn + cn), (cn + dn) are in G.P.

Ans

$\begin{array}{l}\text{\hspace{0.17em}}\mathrm{a},\mathrm{b},\mathrm{c},\mathrm{d}\text{are in G}.\text{P}\\ ⇒\frac{\mathrm{b}}{\mathrm{a}}=\frac{\mathrm{c}}{\mathrm{b}}=\frac{\mathrm{d}}{\mathrm{c}}=\mathrm{r}\left(\mathrm{let}\right)\\ ⇒\mathrm{b}=\mathrm{ar},\mathrm{c}=\mathrm{br}\text{and d}=\text{cr}\\ ⇒\mathrm{b}=\mathrm{ar},\mathrm{c}={\mathrm{ar}}^{2}\text{and d}={\text{ar}}^{\text{3}}\\ \mathrm{Then},\\ \frac{\left({\text{b}}^{\text{n}}+{\text{c}}^{\text{n}}\right)}{\left({\text{a}}^{\text{n}}+{\text{b}}^{\text{n}}\right)}=\frac{\left({\text{a}}^{\text{n}}{\mathrm{r}}^{\mathrm{n}}+{\text{a}}^{\text{n}}{\mathrm{r}}^{2\mathrm{n}}\right)}{\left({\text{a}}^{\text{n}}+{\text{a}}^{\text{n}}{\mathrm{r}}^{\mathrm{n}}\right)}\\ =\frac{{\text{a}}^{\text{n}}{\mathrm{r}}^{\mathrm{n}}\left(1+{\mathrm{r}}^{\mathrm{n}}\right)}{{\text{a}}^{\text{n}}\left(\text{1}+{\mathrm{r}}^{\mathrm{n}}\right)}\\ ={\mathrm{r}}^{\mathrm{n}}\\ \frac{\left({\text{c}}^{\text{n}}+{\text{d}}^{\text{n}}\right)}{\left({\text{b}}^{\text{n}}+{\text{c}}^{\text{n}}\right)}=\frac{\left({\text{a}}^{\text{n}}{\mathrm{r}}^{2\mathrm{n}}+{\text{a}}^{\text{n}}{\mathrm{r}}^{3\mathrm{n}}\right)}{\left({\text{a}}^{\text{n}}{\mathrm{r}}^{\mathrm{n}}+{\text{a}}^{\text{n}}{\mathrm{r}}^{2\mathrm{n}}\right)}\\ =\frac{{\text{a}}^{\text{n}}\text{\hspace{0.17em}}{\mathrm{r}}^{2\mathrm{n}}\left(1+{\mathrm{r}}^{\mathrm{n}}\right)}{{\text{a}}^{\text{n}}\text{\hspace{0.17em}}{\mathrm{r}}^{\mathrm{n}}\left(\text{1}+{\mathrm{r}}^{\mathrm{n}}\right)}\\ ={\mathrm{r}}^{\mathrm{n}}\\ \mathrm{So},\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\frac{\left({\text{b}}^{\text{n}}+{\text{c}}^{\text{n}}\right)}{\left({\text{a}}^{\text{n}}+{\text{b}}^{\text{n}}\right)}=\frac{\left({\text{c}}^{\text{n}}+{\text{d}}^{\text{n}}\right)}{\left({\text{b}}^{\text{n}}+{\text{c}}^{\text{n}}\right)}\\ \mathrm{Therefore},\text{}\left({\text{a}}^{\text{n}}+{\text{b}}^{\text{n}}\right),\left({\text{b}}^{\text{n}}+{\text{c}}^{\text{n}}\right),\left({\text{c}}^{\text{n}}+{\text{d}}^{\text{n}}\right)\text{are G.P.}\end{array}$

Q.76 If a and b are the roots of x2 – 3x + p = 0 and c, d are roots of x2 – 12x + q = 0, where a, b, c, d form a G.P. Prove that (q + p) : (q – p) = 17:15.

Ans

$\begin{array}{l}\mathrm{a}{\text{and b are the roots of x}}^{\text{2}}-\text{3x}+\text{p}=0,\text{then}\\ \text{\hspace{0.17em}sum of zeroes}=-\frac{-3}{1}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{a}+\mathrm{b}=3...\left(\mathrm{i}\right)\\ \text{Product of zeroes}\\ =\frac{\mathrm{p}}{1}\\ \mathrm{ab}=\frac{\mathrm{p}}{1}\\ ⇒\mathrm{ab}=\mathrm{p}\text{\hspace{0.17em}}...\left(\mathrm{ii}\right)\\ \mathrm{c}{\text{and d are the roots of x}}^{\text{2}}-\text{12x}+\text{q}=0,\text{then}\\ \text{sum of zeroes}=-\frac{-12}{1}\\ \text{}\mathrm{c}+\mathrm{d}=12\dots \left(\mathrm{iii}\right)\\ \text{Product of zeroes}\\ =\frac{\mathrm{q}}{1}\\ \text{}\mathrm{cd}=\frac{\mathrm{q}}{1}\\ ⇒\mathrm{cd}=\mathrm{q}\text{\hspace{0.17em}}...\left(\mathrm{iv}\right)\\ \mathrm{Since},\text{a,b,c,d are in G.P.}\\ \text{So,}\frac{\mathrm{b}}{\mathrm{a}}=\frac{\mathrm{c}}{\mathrm{b}}=\frac{\mathrm{d}}{\mathrm{c}}=\mathrm{r}\left(\mathrm{let}\right)\\ ⇒\mathrm{b}=\mathrm{ar},\text{c}=\text{br, d}=\text{cr}\end{array}$ $\begin{array}{l}⇒\mathrm{b}=\mathrm{ar},\text{c}={\text{ar}}^{\text{2}}\text{, d}={\text{ar}}^{\text{3}}\\ \mathrm{From}\text{equation}\left(\mathrm{i}\right),\text{we get}\\ \mathrm{a}+\mathrm{ar}=3\\ \mathrm{a}\left(1+\mathrm{r}\right)=3...\left(\mathrm{v}\right)\\ \mathrm{From}\text{equation}\left(\mathrm{iii}\right),\text{we get}\\ {\mathrm{ar}}^{2}+{\mathrm{ar}}^{3}=12\\ {\mathrm{ar}}^{2}\left(1+\mathrm{r}\right)=12...\left(\mathrm{vi}\right)\\ \mathrm{Dividing}\text{equation}\left(\mathrm{vi}\right)\text{by equation}\left(\mathrm{v}\right),\text{\hspace{0.17em}we get}\\ \frac{{\mathrm{ar}}^{2}\left(1+\mathrm{r}\right)}{\mathrm{a}\left(1+\mathrm{r}\right)}=\frac{12}{3}\\ ⇒\text{\hspace{0.17em}}{\mathrm{r}}^{2}=4\\ ⇒\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{r}=±2\\ \mathrm{Putting}\text{r}=\text{2, in equation}\left(\mathrm{v}\right),\mathrm{we}\text{\hspace{0.17em}\hspace{0.17em}}\mathrm{get}\\ \mathrm{a}\left(1+2\right)=3\\ ⇒\text{\hspace{0.17em}\hspace{0.17em}}\mathrm{a}=\frac{3}{3}=1\\ \mathrm{Putting}\text{r}=-\text{2, in equation}\left(\mathrm{v}\right),\mathrm{we}\text{\hspace{0.17em}\hspace{0.17em}}\mathrm{get}\\ \mathrm{a}\left(1-2\right)=3\\ ⇒\text{\hspace{0.17em}\hspace{0.17em}}\mathrm{a}=\frac{3}{-1}=-3\\ \mathrm{Case}\text{I:}\mathrm{If}\text{a}=\text{1 and r}=\text{2, then}\\ \text{a}=1,\mathrm{b}=1×2,\text{c}=1×{2}^{\text{2}}\text{, d}=1×{2}^{\text{3}}\\ ⇒\mathrm{a}=1,\mathrm{b}=2,\mathrm{c}=4,\mathrm{d}=8\end{array}$ $\begin{array}{l}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\end{array}$ $\begin{array}{l}\text{\hspace{0.17em}}\mathrm{L}.\mathrm{H}.\mathrm{S}.=\frac{\mathrm{q}+\mathrm{p}}{\mathrm{q}-\mathrm{p}}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\frac{\mathrm{cd}+\mathrm{ab}}{\mathrm{cd}-\mathrm{ab}}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\frac{4×8+1×2}{4×8-1×2}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\frac{32+2}{32-2}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\frac{34}{30}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\frac{17}{15}=\mathrm{R}.\mathrm{H}.\mathrm{S}.\\ \mathrm{Case}\text{II:}\mathrm{If}\text{a}=-3\text{and r}=-\text{2, then}\\ \text{a}=-3,\mathrm{b}=-3×-2,\text{c}=-3×{\left(-2\right)}^{\text{2}}\text{, d}=-3×{\left(-2\right)}^{\text{3}}\\ ⇒\mathrm{a}=-3,\mathrm{b}=6,\mathrm{c}=-12,\mathrm{d}=24\\ \mathrm{L}.\mathrm{H}.\mathrm{S}.=\frac{\mathrm{q}+\mathrm{p}}{\mathrm{q}-\mathrm{p}}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\frac{\mathrm{cd}+\mathrm{ab}}{\mathrm{cd}-\mathrm{ab}}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\frac{-12×24+\left(-3\right)×6}{-12×24-\left(-3\right)×6}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}}=\frac{-288-18}{-288+18}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\frac{-306}{-270}\end{array}$ $\begin{array}{l}\text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\end{array}$ $\begin{array}{l}=\frac{17}{15}=\mathrm{R}.\mathrm{H}.\mathrm{S}.\\ \mathrm{Thus},\text{}\left(\mathrm{q}+\mathrm{p}\right):\left(\mathrm{q}-\mathrm{p}\right)=17:15.\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{Hence}\text{proved.}\end{array}$

Q.77

$\begin{array}{l}\text{The ratio of the A.M. and G.M. of two positive numbers a and b, is m : n.}\\ \text{Show that a : b =}\left(\text{m +}\sqrt{{\text{m}}^{\text{2}}-{\text{n}}^{\text{2}}}\right)\text{:}\left(\text{m}-\text{}\sqrt{{\text{m}}^{\text{2}}-{\text{n}}^{\text{2}}}\right)\text{.}\end{array}$

Ans

$\begin{array}{l}\mathrm{We}\text{are given,}\\ \frac{\mathrm{A}.\mathrm{M}.\text{of a and b}}{\mathrm{G}.\mathrm{M}.\text{of a and b}}=\frac{\mathrm{m}}{\mathrm{n}}\\ ⇒\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\frac{\left(\frac{\mathrm{a}+\mathrm{b}}{2}\right)}{\sqrt{\mathrm{ab}}}=\frac{\mathrm{m}}{\mathrm{n}}\\ ⇒\frac{\mathrm{a}+\mathrm{b}}{2\sqrt{\mathrm{ab}}}=\frac{\mathrm{m}}{\mathrm{n}}\\ \mathrm{Apply}\text{componendo and dividendo, we get}\\ ⇒\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\frac{\mathrm{a}+\mathrm{b}+2\sqrt{\mathrm{ab}}}{\mathrm{a}+\mathrm{b}-2\sqrt{\mathrm{ab}}}=\frac{\mathrm{m}+\mathrm{n}}{\mathrm{m}-\mathrm{n}}\\ ⇒\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\frac{{\left(\sqrt{\mathrm{a}}+\sqrt{\mathrm{b}}\right)}^{2}}{{\left(\sqrt{\mathrm{a}}-\sqrt{\mathrm{b}}\right)}^{2}}=\frac{\mathrm{m}+\mathrm{n}}{\mathrm{m}-\mathrm{n}}\\ ⇒\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\frac{\sqrt{\mathrm{a}}+\sqrt{\mathrm{b}}}{\sqrt{\mathrm{a}}-\sqrt{\mathrm{b}}}=\frac{\sqrt{\mathrm{m}+\mathrm{n}}}{\sqrt{\mathrm{m}-\mathrm{n}}}\\ \mathrm{Apply}\text{again componendo and dividendo, we get}\\ \text{\hspace{0.17em}\hspace{0.17em}}\frac{\sqrt{\mathrm{a}}+\sqrt{\mathrm{b}}+\sqrt{\mathrm{a}}-\sqrt{\mathrm{b}}}{\sqrt{\mathrm{a}}+\sqrt{\mathrm{b}}-\sqrt{\mathrm{a}}+\sqrt{\mathrm{b}}}=\frac{\sqrt{\mathrm{m}+\mathrm{n}}+\sqrt{\mathrm{m}-\mathrm{n}}}{\sqrt{\mathrm{m}+\mathrm{n}}-\sqrt{\mathrm{m}-\mathrm{n}}}\end{array}$ $\begin{array}{l}⇒\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\frac{2\sqrt{\mathrm{a}}}{2\sqrt{\mathrm{b}}}=\frac{\sqrt{\mathrm{m}+\mathrm{n}}+\sqrt{\mathrm{m}-\mathrm{n}}}{\sqrt{\mathrm{m}+\mathrm{n}}-\sqrt{\mathrm{m}-\mathrm{n}}}\\ ⇒\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}{\left(\frac{\sqrt{\mathrm{a}}}{\sqrt{\mathrm{b}}}\right)}^{2}=\frac{{\left(\sqrt{\mathrm{m}+\mathrm{n}}+\sqrt{\mathrm{m}-\mathrm{n}}\right)}^{2}}{{\left(\sqrt{\mathrm{m}+\mathrm{n}}-\sqrt{\mathrm{m}-\mathrm{n}}\right)}^{2}}\left[\mathrm{Squarring}\text{both sides}\right]\\ ⇒\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\frac{\mathrm{a}}{\mathrm{b}}=\frac{\mathrm{m}+\mathrm{n}+\mathrm{m}-\mathrm{n}+2\sqrt{\mathrm{m}+\mathrm{n}}\sqrt{\mathrm{m}-\mathrm{n}}}{\mathrm{m}+\mathrm{n}+\mathrm{m}-\mathrm{n}-2\sqrt{\mathrm{m}+\mathrm{n}}\sqrt{\mathrm{m}-\mathrm{n}}}\\ ⇒\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\frac{\mathrm{a}}{\mathrm{b}}=\frac{2\mathrm{m}+2\sqrt{{\mathrm{m}}^{2}-{\mathrm{n}}^{2}}}{2\mathrm{m}-2\sqrt{{\mathrm{m}}^{2}-{\mathrm{n}}^{2}}}\\ ⇒\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\frac{\mathrm{a}}{\mathrm{b}}=\frac{\mathrm{m}+\sqrt{{\mathrm{m}}^{2}-{\mathrm{n}}^{2}}}{\mathrm{m}-\sqrt{{\mathrm{m}}^{2}-{\mathrm{n}}^{2}}}\\ \mathrm{Thus},\text{a:b}=\left(\mathrm{m}+\sqrt{{\mathrm{m}}^{2}-{\mathrm{n}}^{2}}\right):\left(\mathrm{m}-\sqrt{{\mathrm{m}}^{2}-{\mathrm{n}}^{2}}\right).\end{array}$

Q.78

$\begin{array}{l}\text{If a, b, c are in A.P.; b, c, d are in G.P. and}\frac{\text{1}}{\text{c}}\text{,}\frac{\text{1}}{\text{d}}\text{,}\frac{\text{1}}{\text{e}}\text{are in A.P.}\\ \text{prove that a, c, e are in G.P.}\end{array}$

Ans

$\begin{array}{l}\text{a,b,careinA.P.,then}\\ \text{b=}\frac{\text{a+c}}{\text{2}}\\ 2\mathrm{b}\text{}=\text{}\mathrm{a}+\mathrm{c}\text{}...\left(\text{i}\right)\\ \text{b},\text{c},\text{dareinG.P.,then}\\ {\text{c}}^{\text{2}}=\text{bd}\\ \text{d}=\frac{{\text{c}}^{2}}{\text{b}}.\text{}...\left(\mathrm{ii}\right)\\ \frac{1}{\text{c}},\frac{1}{\text{d}},\frac{1}{\text{e}}\text{}\mathrm{are}\text{}\mathrm{in}\text{A}.\text{P}.,\text{then}\\ \frac{2}{\text{d}}=\frac{1}{\text{c}}+\frac{1}{\text{e}}\\ \frac{2}{\text{d}}=\frac{\text{e}+\text{c}}{\mathrm{ce}}\text{}...\left(\mathrm{iii}\right)\\ \mathrm{From}\text{equation}\left(\mathrm{ii}\right)\text{andequation}\left(\mathrm{iii}\right),\text{weget}\\ \text{}\frac{2}{\left(\frac{{\text{c}}^{2}}{\text{b}}\right)}=\frac{\text{e}+\text{c}}{\mathrm{ce}}\\ ⇒\text{}\frac{2\text{b}}{{\text{c}}^{2}}=\frac{\text{e}+\text{c}}{\mathrm{ce}}\\ ⇒\text{}\frac{2\text{b}}{\text{c}}=\frac{\text{e}+\text{c}}{\text{e}}\\ ⇒\text{}\left(\text{a}+\text{c}\right)\text{e}=\mathrm{ce}+{\text{c}}^{2}\text{}\left[\mathrm{From}\text{equation}\left(\text{i}\right)\right]\\ ⇒{\text{ae+ce=ce+c}}^{\text{2}}\\ ⇒{\text{ae=c}}^{\text{2}}\\ ⇒\text{a,c,eareinG.P.}\end{array}$

Q.79 Find the sum of the following series up to n terms:
(i) 5 + 55 + 555 + …
(ii) .6 +. 66 +. 666+…

Ans

$\begin{array}{l}\left(\mathrm{i}\right)\mathrm{Let}\text{\hspace{0.17em}}{\mathrm{S}}_{\mathrm{n}}=\text{5}+\text{55}+\text{555}+\dots \mathrm{}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}}=5\left(1+11+111+1111+...\right)\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\frac{5}{9}\left(9+99+999+9999+...\right)\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\frac{5}{9}\left\{\left(10-1\right)+\left({10}^{2}-1\right)+\left({10}^{3}-1\right)+\left({10}^{4}-1\right)+...\right\}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}}=\frac{5}{9}\left\{10+{10}^{2}+{10}^{3}+{10}^{4}+...-1-1-1-1-...\text{\hspace{0.17em}}\mathrm{n}\text{times}\right\}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}}=\frac{5}{9}\left\{\frac{10\left({10}^{\mathrm{n}}-1\right)}{10-1}-\text{\hspace{0.17em}}\mathrm{n}\right\}\left[âˆµ{\mathrm{S}}_{\mathrm{n}}=\frac{\mathrm{a}\left({\mathrm{r}}^{\mathrm{n}}-1\right)}{\mathrm{r}-1}\right]\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\frac{5}{9}\left\{\frac{10\left({10}^{\mathrm{n}}-1\right)}{9}-\text{\hspace{0.17em}}\mathrm{n}\right\}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}}=\frac{50}{81}\left({10}^{\mathrm{n}}-1\right)-\frac{5\mathrm{n}}{9}\\ \left(\mathrm{ii}\right)\mathrm{Let}\text{\hspace{0.17em}}{\mathrm{S}}_{\mathrm{n}}=0.\text{6}+0.\text{66}+0.\text{666}+\dots \\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}}=\text{6}\left(0.1+0.11+111+1111+...\right)\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\frac{\text{6}}{9}\left(0.9+0.99+0.999+0.9999+...\right)\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\frac{\text{2}}{3}\left\{\left(1-0.1\right)+\left(1-0.01\right)+\left(1-0.001\right)+\left(1-0.0001\right)+...\right\}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}}=\frac{\text{2}}{3}\left\{1+1+1+1+...\text{\hspace{0.17em}}\mathrm{n}\text{times}-\frac{1}{10}-\frac{1}{{10}^{2}}-\frac{1}{{10}^{3}}-\frac{1}{{10}^{4}}-...\right\}\end{array}$ $\begin{array}{l}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}}=\frac{\text{2}}{3}\left\{\mathrm{n}-\frac{\frac{1}{10}\left(1-\frac{1}{{10}^{\mathrm{n}}}\right)}{1-\frac{1}{10}}\right\}\left[âˆµ{\mathrm{S}}_{\mathrm{n}}=\frac{\mathrm{a}\left({\mathrm{r}}^{\mathrm{n}}-1\right)}{\mathrm{r}-1}\right]\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}}=\frac{\text{2}}{3}\mathrm{n}-\frac{\text{2}}{3}×\frac{\left(1-\frac{1}{{10}^{\mathrm{n}}}\right)}{9}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}}=\frac{\text{2}}{3}\mathrm{n}-\frac{\text{2}}{27}×\left(1-\frac{1}{{10}^{\mathrm{n}}}\right)\end{array}$

Q.80 Find the 20th term of the series 2 × 4 + 4 × 6 + 6 × 8 + … + n terms.

Ans

$\begin{array}{l}\text{2}×\text{4}+\text{4}×\text{6}+\text{6}×\text{8}+\text{}...\text{}+\text{n\hspace{0.17em}terms}\\ {\mathrm{T}}_{\mathrm{n}}=\left\{2+\left(\mathrm{n}-1\right)2\right\}\left\{4+\left(\mathrm{n}-1\right)2\right\}\\ =\left(2\mathrm{n}\right)\left(2\mathrm{n}+2\right)\\ =4\mathrm{n}\left(\mathrm{n}+1\right)\\ \mathrm{Putting}\text{n}=20,\text{we get}\\ {\mathrm{T}}_{20}=4\left(20\right)\left(20+1\right)\\ =80×21\\ =1680\\ \mathrm{Thus},{\text{the 20}}^{\text{th}}\text{term of series is 1680.}\end{array}$

Q.81 Find the sum of the first n terms of the series: 3+ 7 + 13 + 21 + 31 +…

Ans

$\begin{array}{l}\mathrm{Let}{\text{S}}_{\text{n}}=3+7+13+21+31+...+{\mathrm{a}}_{\mathrm{n}-1}+{\mathrm{a}}_{\mathrm{n}}\\ \mathrm{and}-\underset{¯}{{\text{S}}_{\text{n}}=\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}-\text{\hspace{0.17em}}3-7-13-21-31-...-{\mathrm{a}}_{\mathrm{n}-1}-{\mathrm{a}}_{\mathrm{n}}}\\ \text{\hspace{0.17em}\hspace{0.17em}}0=3+\left\{4+6+8+10+..\text{\hspace{0.17em}}.\text{\hspace{0.17em}}\left(\mathrm{n}-1\right)\mathrm{terms}\right\}-\text{\hspace{0.17em}}{\mathrm{a}}_{\mathrm{n}}\\ {\mathrm{a}}_{\mathrm{n}}=3+\frac{\mathrm{n}-1}{2}\left\{8+\left(\mathrm{n}-2\right)2\right\}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=3+\left(\mathrm{n}-1\right)\left(4+\mathrm{n}-2\right)\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=3+\left(\mathrm{n}-1\right)\left(\mathrm{n}+2\right)\\ {\mathrm{a}}_{\mathrm{n}}=3+{\mathrm{n}}^{2}+\mathrm{n}-2\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}={\mathrm{n}}^{2}+\mathrm{n}+1\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}{\mathrm{S}}_{\mathrm{n}}=\sum {\mathrm{n}}^{2}+\sum \mathrm{n}+\sum 1\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\frac{\mathrm{n}\left(\mathrm{n}+1\right)\left(2\mathrm{n}+1\right)}{6}+\frac{\mathrm{n}\left(\mathrm{n}+1\right)}{2}+\mathrm{n}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\frac{\mathrm{n}}{6}\left\{\left(\mathrm{n}+1\right)\left(2\mathrm{n}+1\right)+3\left(\mathrm{n}+1\right)+6\right\}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\frac{\mathrm{n}}{6}\left(2{\mathrm{n}}^{2}+3\mathrm{n}+1+3\mathrm{n}+3+6\right)\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\frac{\mathrm{n}}{6}\left(2{\mathrm{n}}^{2}+6\mathrm{n}+10\right)\\ {\mathrm{S}}_{\mathrm{n}}=\frac{\mathrm{n}}{3}\left({\mathrm{n}}^{2}+3\mathrm{n}+5\right)\end{array}$

Q.82

$\begin{array}{l}{\text{If S}}_{\text{1}}{\text{, S}}_{\text{2}}{\text{, S}}_{\text{3}}\text{are the sum of first n natural numbers,their squares and their cubes,respectively,}\\ \text{show that}9{\mathrm{S}}_{2}^{2}={\mathrm{S}}_{3}\left(1+8{\mathrm{S}}_{1}\right).\end{array}$

Ans

$\begin{array}{l}\mathrm{We}\text{are given that}\\ {\text{S}}_{\text{1}}=\sum \mathrm{n}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\frac{\mathrm{n}\left(\mathrm{n}+1\right)}{2}\\ {\text{S}}_{\text{2}}=\sum {\mathrm{n}}^{2}\\ \text{\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}}=\frac{\mathrm{n}\left(\mathrm{n}+1\right)\left(2\mathrm{n}+1\right)}{6}\\ {\text{S}}_{\text{3}}=\sum {\mathrm{n}}^{3}\\ \text{\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}}={\left\{\frac{\mathrm{n}\left(\mathrm{n}+2\right)}{2}\right\}}^{2}\\ \mathrm{R}.\mathrm{H}.\mathrm{S}.={\mathrm{S}}_{3}\left(1+8{\mathrm{S}}_{1}\right)\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}={\left\{\frac{\mathrm{n}\left(\mathrm{n}+2\right)}{2}\right\}}^{2}\left\{1+8×\frac{\mathrm{n}\left(\mathrm{n}+1\right)}{2}\right\}\\ \text{\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}}=\frac{{\mathrm{n}}^{2}{\left(\mathrm{n}+2\right)}^{2}}{4}\left\{1+4\mathrm{n}\left(\mathrm{n}+1\right)\right\}\\ \text{\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}}=\frac{{\mathrm{n}}^{2}{\left(\mathrm{n}+2\right)}^{2}}{4}\left(1+4{\mathrm{n}}^{2}+4\mathrm{n}\right)\\ \text{\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}}=\frac{{\mathrm{n}}^{2}{\left(\mathrm{n}+2\right)}^{2}}{4}\left(4{\mathrm{n}}^{2}+4\mathrm{n}+1\right)\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\frac{{\mathrm{n}}^{2}{\left(\mathrm{n}+2\right)}^{2}{\left(2\mathrm{n}+1\right)}^{2}}{4}\\ \text{\hspace{0.17em} \hspace{0.17em}\hspace{0.17em}}={\left\{3×\frac{\mathrm{n}\left(\mathrm{n}+2\right)\left(2\mathrm{n}+1\right)}{6}\right\}}^{2}\end{array}$ $\begin{array}{l}\text{\hspace{0.17em} \hspace{0.17em}\hspace{0.17em}}=9{\left\{\frac{\mathrm{n}\left(\mathrm{n}+2\right)\left(2\mathrm{n}+1\right)}{6}\right\}}^{2}\\ =9\text{\hspace{0.17em}}{\mathrm{S}}_{2}^{2}=\mathrm{L}.\mathrm{H}.\mathrm{S}.\\ \mathrm{This}\text{was to be proved.}\end{array}$

Q.83

$\begin{array}{l}\mathbf{\text{Find the sum of the following series up to n terms:}}\\ \frac{{\mathbf{\text{1}}}^{\mathbf{\text{3}}}}{\mathbf{\text{1}}}\mathbf{\text{+}}\frac{{\mathbf{\text{1}}}^{\mathbf{\text{3}}}{\mathbf{\text{+2}}}^{\mathbf{\text{3}}}}{\mathbf{\text{1+3}}}\mathbf{\text{+}}\frac{{\mathbf{\text{1}}}^{\mathbf{\text{3}}}{\mathbf{\text{+2}}}^{\mathbf{\text{3}}}{\mathbf{\text{+3}}}^{\mathbf{\text{3}}}}{\mathbf{\text{1+3+5}}}\mathbf{\text{+}}\mathbf{.}\mathbf{..}\end{array}$

Ans

$\begin{array}{l}\frac{{1}^{3}}{1}+\frac{{1}^{3}+{2}^{3}}{1+3}+\frac{{1}^{3}+{2}^{3}+{3}^{3}}{1+3+5}+...\\ {\mathrm{T}}_{\mathrm{n}}=\frac{{1}^{3}+{2}^{3}+{3}^{3}+...+{\mathrm{n}}^{3}}{1+3+5+...}\\ =\frac{\sum {\mathrm{n}}^{3}}{\frac{\mathrm{n}}{2}\left\{2×1+\left(\mathrm{n}-1\right)2\right\}}\\ =\frac{{\left\{\frac{\mathrm{n}\left(\mathrm{n}+1\right)}{2}\right\}}^{2}}{\frac{\mathrm{n}}{2}\left(2+2\mathrm{n}-2\right)}\\ =\frac{{\mathrm{n}}^{2}{\left(\mathrm{n}+1\right)}^{2}}{4{\mathrm{n}}^{2}}\\ =\frac{1}{4}\left({\mathrm{n}}^{2}+2\mathrm{n}+1\right)\end{array}$ $\begin{array}{l}{\mathrm{S}}_{\mathrm{n}}=\frac{1}{4}\left(\sum {\mathrm{n}}^{2}+2\sum \mathrm{n}+\sum 1\right)\\ =\frac{1}{4}\left[\frac{\mathrm{n}\left(\mathrm{n}+1\right)\left(2\mathrm{n}+1\right)}{6}+\frac{2\mathrm{n}\left(\mathrm{n}+1\right)}{2}+\mathrm{n}\right]\\ =\frac{1}{4}\left\{\frac{\mathrm{n}\left(\mathrm{n}+1\right)\left(2\mathrm{n}+1\right)}{6}+\mathrm{n}+1+1\right\}\\ =\frac{\mathrm{n}}{4}\left(\frac{2{\mathrm{n}}^{2}+3\mathrm{n}+1+6\mathrm{n}+12}{6}\right)\\ =\frac{\mathrm{n}}{4}\left(\frac{2{\mathrm{n}}^{2}+9\mathrm{n}+13}{6}\right)\\ =\frac{\mathrm{n}}{24}\left(2{\mathrm{n}}^{2}+9\mathrm{n}+13\right)\end{array}$

Q.84

$\begin{array}{l}\mathrm{Show}\mathrm{that}\\ \frac{1×{2}^{2}+2×{3}^{2}+...+\mathrm{n}×{\left(\mathrm{n}+1\right)}^{2}}{{1}^{2}×2+{2}^{2}×3+...+{\mathrm{n}}^{2}×\left(\mathrm{n}+1\right)}=\frac{3\mathrm{n}+5}{3\mathrm{n}+1}\end{array}$

Ans

$\begin{array}{l}\mathrm{L}.\mathrm{H}.\mathrm{S}.=\frac{1×{2}^{2}+2×{3}^{2}+...+\mathrm{n}×{\left(\mathrm{n}+1\right)}^{2}}{{1}^{2}×2+{2}^{2}×3+...+{\mathrm{n}}^{2}×\left(\mathrm{n}+1\right)}\\ \text{\hspace{0.17em}\hspace{0.17em}}=\frac{\sum \left\{\mathrm{n}×{\left(\mathrm{n}+1\right)}^{2}\right\}}{\sum \left\{{\mathrm{n}}^{2}×\left(\mathrm{n}+1\right)\right\}}\\ \text{\hspace{0.17em}\hspace{0.17em}}=\frac{\sum \left({\mathrm{n}}^{3}+2{\mathrm{n}}^{2}+\mathrm{n}\right)}{\sum \left({\mathrm{n}}^{3}+{\mathrm{n}}^{2}\right)}\end{array}$ $\begin{array}{l}\text{\hspace{0.17em}\hspace{0.17em}}=\frac{\sum {\mathrm{n}}^{3}+2\sum {\mathrm{n}}^{2}+\sum \mathrm{n}}{\sum {\mathrm{n}}^{3}+\sum {\mathrm{n}}^{2}}\\ \text{\hspace{0.17em}\hspace{0.17em}}=\frac{{\left\{\frac{\mathrm{n}\left(\mathrm{n}+1\right)}{2}\right\}}^{2}+2\left\{\frac{\mathrm{n}\left(\mathrm{n}+1\right)\left(2\mathrm{n}+1\right)}{6}\right\}+\frac{\mathrm{n}\left(\mathrm{n}+1\right)}{2}}{{\left\{\frac{\mathrm{n}\left(\mathrm{n}+1\right)}{2}\right\}}^{2}+\frac{\mathrm{n}\left(\mathrm{n}+1\right)\left(2\mathrm{n}+1\right)}{6}}\\ \text{\hspace{0.17em}\hspace{0.17em}}=\frac{\frac{\mathrm{n}\left(\mathrm{n}+1\right)}{2}\left\{\frac{\mathrm{n}\left(\mathrm{n}+1\right)}{2}+2×\frac{2\mathrm{n}+1}{3}+1\right\}}{\frac{\mathrm{n}\left(\mathrm{n}+1\right)}{2}\left\{\frac{\mathrm{n}\left(\mathrm{n}+1\right)}{2}+\frac{2\mathrm{n}+1}{3}\right\}}\\ \text{\hspace{0.17em}\hspace{0.17em}}=\frac{\left(\frac{3{\mathrm{n}}^{2}+3\mathrm{n}+8\mathrm{n}+4+6}{6}\right)}{\left(\frac{3{\mathrm{n}}^{2}+3\mathrm{n}+4\mathrm{n}+2}{6}\right)}\\ =\frac{3{\mathrm{n}}^{2}+11\mathrm{n}+10}{3{\mathrm{n}}^{2}+7\mathrm{n}+2}\\ =\frac{\left(\mathrm{n}+2\right)\left(3\mathrm{n}+5\right)}{\left(\mathrm{n}+2\right)\left(3\mathrm{n}+1\right)}\end{array}$ $\begin{array}{l}\text{}\end{array}$ $\begin{array}{l}=\frac{\left(3\mathrm{n}+5\right)}{\left(3\mathrm{n}+1\right)}=\mathrm{R}.\mathrm{H}.\mathrm{S}.\\ \mathrm{This}\text{was to be proved.}\end{array}$

Q.85 150 workers were engaged to finish a job in a certain number of days. 4 workers dropped out on second day, 4 more workers dropped out on third day and so on. It took 8 more days to finish the work. Find the number of days in which the work was completed.

Ans

$\begin{array}{l}\mathrm{Let}\text{150 workers finish a job in n days.}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}Number of workers started the job}=150\\ \text{Number of workers left the job per day}=4\\ \mathrm{The}\text{series of workers remain on the job every day}\\ 150,146,144,140,...\\ \mathrm{Then},\\ \mathrm{Number}\text{of workers finished job in a day}=\mathrm{Total}\text{}\mathrm{Number}\text{of}\\ \text{reducing workers finished job in}\left(\mathrm{n}+8\right)\mathrm{days}\\ ⇒\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}150\text{\hspace{0.17em}}\mathrm{n}=150+146+144\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}+...\left(\mathrm{n}+8\right)\text{terms}\\ ⇒\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}150\text{\hspace{0.17em}}\mathrm{n}=\frac{\mathrm{n}+8}{2}\left\{\begin{array}{l}2×150+\\ \left(\mathrm{n}+8-1\right)\left(-4\right)\end{array}\right\}\\ ⇒\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}300\text{\hspace{0.17em}}\mathrm{n}=\left(\mathrm{n}+8\right)\left\{300-4\left(\mathrm{n}+7\right)\right\}\\ ⇒\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}300\text{\hspace{0.17em}}\mathrm{n}=\left(\mathrm{n}+8\right)\left(300-4\mathrm{n}-28\right)\end{array}$ $\begin{array}{l}⇒\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}300\text{\hspace{0.17em}}\mathrm{n}=4\left(\mathrm{n}+8\right)\left(68-\mathrm{n}\right)\\ ⇒\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}75\text{\hspace{0.17em}}\mathrm{n}=-\left({\mathrm{n}}^{2}-60\text{\hspace{0.17em}}\mathrm{n}-544\right)\\ ⇒\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}{\mathrm{n}}^{2}-60\text{\hspace{0.17em}}\mathrm{n}-544+75\text{\hspace{0.17em}}\mathrm{n}=0\end{array}$ $\begin{array}{l}⇒\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}{\mathrm{n}}^{2}+15\text{\hspace{0.17em}}\mathrm{n}-544=0\\ ⇒\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\left(\mathrm{n}+32\right)\left(\mathrm{n}-17\right)=0\\ ⇒\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{n}=17,-32\left(\mathrm{Negative}\right)\\ \mathrm{Thus},\text{the number of days required to complete work by 150}\\ \text{workers is 17 days.Therefore, number of days required to finish}\\ \text{the job by reducing number of workers}=18+7\\ =25\mathrm{days}\end{array}$