# 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.

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### 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 theÂ  Extramarksâ€™ 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.Â

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 â€˜nâ€™ terms of the series.Â

Â

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, â€¦â€¦â€¦â€¦, 1024
• 3, .3, .33, .333, .3333â€¦â€¦and so on.

The different elements in a sequence are called terms, for example (a1, a2, a3, a4,â€¦â€¦ 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.Â

• Finite sequenceÂ

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

Fibonacci sequence :

Example-Â

A1 = A2 = 1,Â

A3 = A1 + A2,Â

An = An-2 + An-1,

Â n > 2

Series

In this section, we will learn about the series.

There is an example given below that shows the series:

a1 +Â  a2 +Â  a3 +Â  a4 + â€¦â€¦ an + â€¦..

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 â€˜âˆ‘â€™.

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 +Â  a2 +Â  a3 +Â  a4 + â€¦â€¦ an + â€¦..

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 + 3dâ€¦a + (n + 1)d

Formulas,

• The last number of A.P,

l = a + (n – 1)d

• The sum of n numbers in A.P,Â

SnÂ  = n/2 [2a + (n – 1)d]

Â Â Â Â Â Â Â Or

SnÂ  = 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.Â

There is a series given below which follows arithmetic progression:

a1 +Â  a2 +Â  a3 +Â  a4 + â€¦â€¦ an + â€¦..

Â ak+1 / ak = r (constant), for k â‰¥ 1

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

The series given below follows a Geometric progression:

a, ar, ar2, ar3,â€¦..

Formulas,

• General â€˜nâ€™ term of a G.P,

anÂ  =Â  arn-1

• The sum of n numbers in G.P,Â

SnÂ  = a(1-rn) / (1-r)Â Â

Â Â Â Â Â Â Â Or

SnÂ  = a(rn-1) / (r-1)

Geometric mean

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

Relationship between A.M and G.M

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

A.M = (a + b) /2

G.M = âˆša.b

Then, A.M – G.M = (a + b) /2 – âˆša.b = (âˆša – âˆšb)2 / 2 â‰¥ 0.Â

This is the relationship between A.M â‰¥ 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.Â

• 1 + 2 + 3 + 4 +â€¦â€¦.+n (sum of first â€˜nâ€™ numbers)

Sn = n(n + 1) / 2Â

• 12 + 22 + 32 + 42 +â€¦â€¦.+n2 (sum of first square â€˜nâ€™ numbers)

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

• 13 + 23 + 33 + 43 +â€¦â€¦.+n3(sum of first cube â€˜nâ€™ 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 Extramarksâ€™ 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 theÂ  link belowÂ  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 ExerciseÂ Â

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

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

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.Â

The key features are as follows:Â

• The entire chapter has been summarised in a point-wise manner in our NCERT solutions.Â
• 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

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

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

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

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

Ans

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

Q.4

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

Ans

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

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

Ans

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

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

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

Q.7

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

Ans

1.

$\begin{array}{l}\text{Given:}\\ {\text{a}}_{\text{n}}=\text{4n}â€“\text{3}\\ \mathrm{Substituting}\text{n=17 and 24 respectively, we get}\\ {\text{a}}_{\text{17}}=\text{4}\left(17\right)âˆ’\text{3}\\ \text{â€‰â€‰â€‰â€‰â€‰â€‰â€‰}=\text{68}âˆ’\text{3}\\ \text{â€‰â€‰â€‰â€‰â€‰â€‰â€‰}=\text{65}\\ {\text{a}}_{\text{24}}=\text{4}\left(24\right)âˆ’\text{3}\\ \text{â€‰â€‰â€‰â€‰â€‰â€‰â€‰}=96âˆ’3\\ \text{â€‰â€‰â€‰â€‰â€‰â€‰â€‰}=93\end{array}$

2.

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

3.

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

Q.8

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

Ans

1.

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

2.

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

3.

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

Q.9

$\begin{array}{l}{\mathbf{\text{The Fibonacci sequence is defined by 1=a}}}_{\mathbf{\text{1}}}{\mathbf{\text{=a}}}_{\mathbf{\text{2}}}{\mathbf{\text{and a}}}_{\mathbf{\text{n}}}{\mathbf{\text{=a}}}_{\mathbf{\text{n}}\mathbf{âˆ’}\mathbf{\text{1}}}{\mathbf{\text{+a}}}_{\mathbf{\text{n}}\mathbf{âˆ’}\mathbf{\text{2}}}\mathbf{\text{,â€‰â€‰n>2.}}\\ \mathbf{\text{Find}}\frac{{\mathbf{\text{a}}}_{\mathbf{\text{n+1}}}}{{\mathbf{\text{a}}}_{\mathbf{\text{n}}}}\mathbf{\text{, for n=1,2,3,4,5.}}\end{array}$

Ans

$\begin{array}{l}\text{TheFibonaccisequenceisdefinedby}\\ {\text{1=a}}_{\text{1}}{\text{=a}}_{\text{2}}{\text{anda}}_{\text{n}}{\text{=a}}_{\text{n}âˆ’\text{1}}{\text{+a}}_{\text{n}âˆ’\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}âˆ’\text{1}}}{{\text{a}}_{\text{n-1}}{\text{+a}}_{\text{n}âˆ’\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}âˆ’\text{1}}}{{\text{a}}_{\text{3}âˆ’\text{1}}{\text{+a}}_{\text{3}âˆ’\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

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

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

Ans

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

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 â€“112.

Ans

$\begin{array}{l}\text{â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰}\mathrm{First}\text{term of A.P.}=\text{2}\\ \text{Let common difference}=\mathrm{d}\\ \mathrm{According}\text{to given condition:}\\ \text{Sum of first 5 terms}=\frac{1}{4}\left(\mathrm{Sum}\text{of next 5 terms}\right)\\ \text{â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰}{\mathrm{S}}_{5}=\frac{1}{4}\left({\mathrm{S}}_{10}âˆ’{\mathrm{S}}_{5}\right)\\ â‡’\text{â€‰â€‰â€‰â€‰â€‰â€‰}4{\mathrm{S}}_{5}={\mathrm{S}}_{10}âˆ’{\mathrm{S}}_{5}\\ â‡’\text{â€‰â€‰â€‰â€‰â€‰â€‰}5{\mathrm{S}}_{5}={\mathrm{S}}_{10}\end{array}$ $\begin{array}{l}â‡’5\left[\frac{5}{2}\left\{2Ã—2+\left(5âˆ’1\right)\mathrm{d}\right\}\right]=\left[\frac{10}{2}\left\{2Ã—2+\left(10âˆ’1\right)\mathrm{d}\right\}\right]\\ \left[Ã¢Ë†Âµ{\mathrm{S}}_{\mathrm{n}}=\frac{\mathrm{n}}{2}\left\{2\mathrm{a}+\left(\mathrm{n}âˆ’1\right)\mathrm{d}\right\}\right]\\ â‡’\text{â€‰â€‰â€‰â€‰â€‰}\frac{25}{2}\left(4+4\mathrm{d}\right)=5\left(4+9\mathrm{d}\right)\\ â‡’\text{â€‰â€‰â€‰â€‰â€‰â€‰}25\left(2+2\mathrm{d}\right)=5\left(4+9\mathrm{d}\right)\\ â‡’\text{â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰}50+50\mathrm{d}=20+45\mathrm{d}\\ â‡’\text{â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰}5\mathrm{d}=20âˆ’50\\ â‡’\text{â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰}\mathrm{d}=âˆ’\frac{30}{5}\\ \text{â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰}=âˆ’6\\ \mathrm{Now},\text{sum of 20 terms.}\\ {\text{â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰T}}_{\text{20}}=2+\left(20âˆ’1\right)Ã—âˆ’6\left[Ã¢Ë†Âµ{\mathrm{a}}_{\mathrm{n}}=\mathrm{a}+\left(\mathrm{n}âˆ’1\right)\mathrm{d}\right]\\ \text{â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰}=2âˆ’114\\ \text{â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰}=âˆ’112\\ \mathrm{Thus},{\text{the 20}}^{\text{th}}\text{term is}âˆ’\text{112.}\end{array}$

Q.13

$\begin{array}{l}\mathbf{\text{How many terms of the A.P.â€‰â€‰}}\mathbf{âˆ’}\mathbf{\text{6,}}\mathbf{âˆ’}\frac{\mathbf{\text{11}}}{\mathbf{\text{2}}}\mathbf{\text{,}}\mathbf{âˆ’}\mathbf{\text{5,}}\mathbf{.}\mathbf{..}\mathbf{\text{â€‰ are needed to give}}\\ \mathbf{\text{the sum}}\mathbf{âˆ’}\mathbf{\text{25?}}\end{array}$

Ans

$\begin{array}{l}\mathrm{The}\text{given A.P. is}âˆ’6,âˆ’\frac{11}{2},âˆ’5,...\text{â€‰}\\ \mathrm{Let}\text{the sum of n terms of A.P. is}âˆ’\text{25.Then,}\end{array}$ $\begin{array}{l}{\mathrm{S}}_{\mathrm{n}}=\frac{\mathrm{n}}{2}\left\{2\mathrm{a}+\left(\mathrm{n}âˆ’1\right)\mathrm{d}\right\}\\ \mathrm{a}=âˆ’6\\ \mathrm{d}=âˆ’\frac{11}{2}âˆ’\left(âˆ’6\right)\\ \text{â€‰â€‰â€‰â€‰â€‰â€‰}=âˆ’\frac{11}{2}+6=\frac{1}{2}\\ âˆ’25=\frac{\mathrm{n}}{2}\left\{2Ã—âˆ’6+\left(\mathrm{n}âˆ’1\right)Ã—\frac{1}{2}\right\}\left[Ã¢Ë†Âµ{\mathrm{S}}_{\mathrm{n}}=\frac{\mathrm{n}}{2}\left\{2\mathrm{a}+\left(\mathrm{n}âˆ’1\right)\mathrm{d}\right\}\right]\\ âˆ’25=\frac{\mathrm{n}}{2}\left\{âˆ’12+\left(\mathrm{n}âˆ’1\right)Ã—\frac{1}{2}\right\}\\ âˆ’100=\mathrm{n}\left(âˆ’24+\mathrm{n}âˆ’1\right)\\ âˆ’100=âˆ’25\text{â€‰}\mathrm{n}+{\mathrm{n}}^{2}\\ {\mathrm{n}}^{2}âˆ’25\text{â€‰}\mathrm{n}+100=0\\ \left(\mathrm{n}âˆ’20\right)\left(\mathrm{n}âˆ’5\right)=0\\ â‡’\text{â€‰ â€‰â€‰â€‰â€‰â€‰â€‰}\mathrm{n}=20,\text{â€‰â€‰}5\\ \mathrm{Thus},\text{the number of terms of A.P. is 5 or 20.}\end{array}$

Q.14

Ans

$\begin{array}{l}\text{â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰}\mathrm{Let}\text{first term of A.P.}=\mathrm{a}\\ \mathrm{Let}\text{common difference of A.P.}=\mathrm{d}\\ \mathrm{Since},{\text{T}}_{\text{p}}=\frac{1}{\mathrm{q}}\end{array}$ $\begin{array}{l}â‡’\mathrm{a}+\left(\mathrm{p}âˆ’1\right)\mathrm{d}=\frac{1}{\mathrm{q}}...\left(\mathrm{i}\right)\\ \mathrm{And},{\text{T}}_{\text{q}}=\frac{1}{\mathrm{p}}\\ â‡’\mathrm{a}+\left(\mathrm{q}âˆ’1\right)\mathrm{d}=\frac{1}{\mathrm{p}}...\left(\mathrm{ii}\right)\\ \mathrm{Subtracting}\text{equation}\left(\mathrm{ii}\right)\text{from equation}\left(\mathrm{i}\right),\text{we get}\\ \text{â€‰â€‰â€‰}\left(\mathrm{p}âˆ’1âˆ’\mathrm{q}+1\right)\mathrm{d}=\frac{1}{\mathrm{q}}âˆ’\frac{1}{\mathrm{p}}\\ â‡’\text{â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰}\left(\mathrm{p}âˆ’\mathrm{q}\right)\mathrm{d}=\frac{\mathrm{p}âˆ’\mathrm{q}}{\mathrm{pq}}\\ â‡’\text{â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰}\mathrm{d}=\frac{1}{\mathrm{pq}}\\ \mathrm{Substituting}\text{value of d in equation}\left(\mathrm{i}\right),\text{â€‰}\mathrm{we}\text{â€‰}\mathrm{get}\\ \text{â€‰â€‰â€‰â€‰â€‰â€‰}\mathrm{a}+\left(\mathrm{p}âˆ’1\right)\frac{1}{\mathrm{pq}}=\frac{1}{\mathrm{q}}\\ â‡’\text{â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰}\mathrm{a}+\frac{\mathrm{p}}{\mathrm{pq}}âˆ’\frac{1}{\mathrm{pq}}=\frac{1}{\mathrm{q}}\\ â‡’\text{â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰}\mathrm{a}=\frac{1}{\mathrm{pq}}\\ \mathrm{Now},\text{â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰}{\mathrm{S}}_{\mathrm{pq}}=\frac{\mathrm{pq}}{2}\left\{2Ã—\frac{1}{\mathrm{pq}}+\left(\mathrm{pq}âˆ’1\right)\frac{1}{\mathrm{pq}}\right\}\\ \left[Ã¢Ë†Âµ{\mathrm{S}}_{\mathrm{n}}=\frac{\mathrm{n}}{2}\left\{2\mathrm{a}+\left(\mathrm{n}âˆ’1\right)\mathrm{d}\right\}\right]\end{array}$ $\begin{array}{l}\text{â€‰}\text{â€‰}\text{â€‰}\text{â€‰}\text{â€‰}\text{â€‰}\text{}\text{}\text{}\text{}\text{â€‰}\text{â€‰}\text{â€‰}\text{â€‰}\text{â€‰}\end{array}$ $\begin{array}{l}\text{â€‰}{\mathrm{S}}_{\mathrm{pq}}=\frac{\mathrm{pq}}{2}\left\{2Ã—\frac{1}{\mathrm{pq}}+\left(\mathrm{pq}âˆ’1\right)\frac{1}{\mathrm{pq}}\right\}\\ \text{â€‰â€‰â€‰}=\frac{\mathrm{pq}}{2}\left(\frac{2}{\mathrm{pq}}+1âˆ’\frac{1}{\mathrm{pq}}\right)\\ \text{â€‰â€‰â€‰}=\frac{\mathrm{pq}}{2}\left(\frac{1}{\mathrm{pq}}+1\right)\\ \text{â€‰â€‰â€‰}=\frac{1}{2}\left(1+\mathrm{pq}\right)\\ \mathrm{Thus},\text{the sum of pq terms is}\frac{1}{2}\left(\mathrm{pq}+1\right).\end{array}$

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

Ans

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

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

Ans

$\begin{array}{l}{\mathrm{k}}^{\mathrm{th}}\text{term of A.P.}=\text{5k+1}\\ â‡’\text{â€‰}{\mathrm{T}}_{\mathrm{k}}=\text{5k+1}\\ \mathrm{Substituting}\text{k}=1,2,3,...\text{respectively.}\\ {\text{â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰T}}_{\text{1}}=5\left(1\right)+1\\ \text{â€‰â€‰â€‰â€‰}=6\\ {\text{â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰T}}_{\text{2}}=5\left(2\right)+1\\ \text{â€‰â€‰â€‰â€‰}=11\\ {\text{â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰T}}_{\text{3}}=5\left(3\right)+1\\ \text{â€‰â€‰â€‰â€‰}=16\\ \mathrm{Common}\text{difference}={\mathrm{T}}_{2}âˆ’{\mathrm{T}}_{1}\\ \text{â€‰â€‰â€‰â€‰}=11âˆ’6\\ \text{â€‰â€‰â€‰â€‰}=5\end{array}$ $\begin{array}{l}\mathrm{Sum}{\text{of n terms,S}}_{\text{n}}=\frac{\mathrm{n}}{2}\left\{2\mathrm{a}+\left(\mathrm{n}âˆ’1\right)\mathrm{d}\right\}\\ \text{â€‰â€‰â€‰â€‰}=\frac{\mathrm{n}}{2}\left\{2Ã—6+\left(\mathrm{n}âˆ’1\right)5\right\}\\ \text{â€‰â€‰â€‰â€‰}=\frac{\mathrm{n}}{2}\left(12+5\mathrm{n}âˆ’5\right)\\ \text{â€‰â€‰â€‰â€‰}=\frac{\mathrm{n}}{2}\left(7+5\mathrm{n}\right)\\ \text{â€‰â€‰â€‰â€‰}=\frac{\mathrm{n}}{2}\left(5\mathrm{n}+7\right)\\ \mathrm{Thus},\text{the sum of n terms is}\frac{\mathrm{n}}{2}\left(5\mathrm{n}+7\right).\end{array}$

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

$\begin{array}{l}\mathrm{Given},\text{â€‰}{\mathrm{S}}_{\mathrm{n}}=\left(\mathrm{pn}+{\mathrm{qn}}^{2}\right)\\ \mathrm{Putting}\text{n}=1,2,3\text{â€‰}\mathrm{respectively},\mathrm{we}\text{â€‰}\mathrm{get}\\ \text{â€‰â€‰â€‰â€‰â€‰}{\mathrm{S}}_{1}=\left(\mathrm{p}Ã—1+\mathrm{q}Ã—{1}^{2}\right)\\ \text{â€‰}=\mathrm{p}+\mathrm{q}\\ \text{â€‰â€‰â€‰â€‰â€‰}{\mathrm{S}}_{2}=\left(\mathrm{p}Ã—2+\mathrm{q}Ã—{2}^{2}\right)\\ \text{â€‰}=2\text{â€‰}\mathrm{p}+4\text{â€‰}\mathrm{q}\\ \text{â€‰â€‰â€‰â€‰â€‰}{\mathrm{S}}_{3}=\left(\mathrm{p}Ã—3+\mathrm{q}Ã—{3}^{2}\right)\\ \text{â€‰}=3\text{â€‰}\mathrm{p}+9\text{â€‰}\mathrm{q}\\ \text{â€‰â€‰â€‰â€‰â€‰â€‰}{\mathrm{T}}_{1}={\mathrm{S}}_{2}âˆ’{\mathrm{S}}_{1}\\ \text{â€‰}=\left(2\text{â€‰}\mathrm{p}+4\text{â€‰}\mathrm{q}\right)âˆ’\left(\mathrm{p}+\mathrm{q}\right)\end{array}$ $\begin{array}{l}\text{â€‰}=\mathrm{p}+3\mathrm{q}\\ {\mathrm{T}}_{2}={\mathrm{S}}_{3}âˆ’{\mathrm{S}}_{2}\\ \text{â€‰}=\left(3\text{â€‰}\mathrm{p}+9\text{â€‰}\mathrm{q}\right)âˆ’\left(2\text{â€‰}\mathrm{p}+4\text{â€‰}\mathrm{q}\right)\\ \text{â€‰}=\mathrm{p}+5\text{â€‰}\mathrm{q}\\ \mathrm{Common}\text{difference,}\\ \text{d}={\mathrm{T}}_{2}âˆ’{\mathrm{T}}_{1}\\ \text{â€‰}=\left(\mathrm{p}+5\text{â€‰}\mathrm{q}\right)âˆ’\left(\mathrm{p}+3\mathrm{q}\right)\\ \text{â€‰}=2\mathrm{q}\\ \mathrm{Thus},\text{common difference is 2q.}\end{array}$

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

$\begin{array}{l}\mathrm{Since},\\ \frac{\mathrm{Sum}\text{of n terms of an A.P.}}{\mathrm{Sum}\text{of n terms of another A.P.}}=\frac{\text{5n}+\text{4}}{\text{9n}+\text{6}}\\ â‡’\text{â€‰â€‰â€‰â€‰}\frac{\frac{\mathrm{n}}{2}\left\{2\mathrm{a}+\left(\mathrm{n}âˆ’1\right)\mathrm{d}\right\}}{\frac{\mathrm{n}}{2}\left\{2\mathrm{A}+\left(\mathrm{n}âˆ’1\right)\mathrm{D}\right\}}=\frac{\text{5n}+\text{4}}{\text{9n}+\text{6}}\\ â‡’\text{â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰}\frac{\left\{\mathrm{a}+\left(\frac{\mathrm{n}âˆ’1}{2}\right)\mathrm{d}\right\}}{\left\{\mathrm{A}+\left(\frac{\mathrm{n}âˆ’1}{2}\right)\mathrm{D}\right\}}=\frac{\text{5n}+\text{4}}{\text{9n}+\text{6}}\text{â€‰ â€‰â€‰}...\left(\mathrm{i}\right)\\ \mathrm{Since},{\text{T}}_{\text{18}}=\text{a}+\text{17d. So, Substituting}\frac{\mathrm{n}âˆ’1}{2}=17\end{array}$ $\begin{array}{l}\mathrm{or}\text{â€‰}\mathrm{n}=35\text{in equation}\left(\mathrm{i}\right),\text{we get}\\ \frac{\left\{\mathrm{a}+17\mathrm{d}\right\}}{\left\{\mathrm{A}+17\mathrm{D}\right\}}=\frac{\text{5}Ã—\text{35}+\text{4}}{\text{9}Ã—\text{35}+\text{6}}\\ \frac{{\text{18}}^{\text{th}}\text{term of an A.P.}}{{\text{18}}^{\text{th}}\text{term of another A.P.}}=\frac{\text{175}+\text{4}}{\text{315}+\text{6}}\\ =\frac{179}{321}\\ \mathrm{Thus},{\text{the ratio of 18}}^{\text{th}}\text{terms of two A.P. is 179:321.}\end{array}$

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

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

Q.20

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

Ans

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

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 â€“ 1) : (2n â€“ 1).

Ans

$\begin{array}{l}\mathrm{Let}\text{first term of an A.P.}=\text{a}\\ \text{and commong difference of A.P.}=\mathrm{d}\\ \mathrm{It}\text{is given that:}\\ {\text{S}}_{\text{m}}{\text{:S}}_{\text{n}}={\mathrm{m}}^{2}:{\mathrm{n}}^{2}\\ â‡’\text{â€‰}\frac{{\mathrm{S}}_{\mathrm{m}}}{{\mathrm{S}}_{\mathrm{n}}}=\frac{{\mathrm{m}}^{2}}{{\mathrm{n}}^{2}}\end{array}$ $\begin{array}{l}â‡’\frac{\frac{\mathrm{m}}{2}\left\{2\mathrm{a}+\left(\mathrm{m}âˆ’1\right)\mathrm{d}\right\}}{\frac{\mathrm{n}}{2}\left\{2\mathrm{a}+\left(\mathrm{n}âˆ’1\right)\mathrm{d}\right\}}=\frac{{\mathrm{m}}^{2}}{{\mathrm{n}}^{2}}\\ â‡’\frac{\left\{2\mathrm{a}+\left(\mathrm{m}âˆ’1\right)\mathrm{d}\right\}}{\left\{2\mathrm{a}+\left(\mathrm{n}âˆ’1\right)\mathrm{d}\right\}}=\frac{\mathrm{m}}{\mathrm{n}}\\ â‡’2\mathrm{an}+\left(\mathrm{mn}âˆ’\mathrm{n}\right)\mathrm{d}=2\mathrm{am}+\left(\mathrm{mn}âˆ’\mathrm{m}\right)\mathrm{d}\\ â‡’2\mathrm{an}âˆ’2\mathrm{am}=\left(\mathrm{mn}âˆ’\mathrm{m}\right)\mathrm{d}âˆ’\left(\mathrm{mn}âˆ’\mathrm{n}\right)\mathrm{d}\\ â‡’\text{}2\mathrm{a}\left(\mathrm{n}âˆ’\mathrm{m}\right)=\mathrm{d}\left(\mathrm{mn}âˆ’\mathrm{m}âˆ’\mathrm{mn}+\mathrm{n}\right)\\ â‡’\text{}2\mathrm{a}\left(\mathrm{n}âˆ’\mathrm{m}\right)=\mathrm{d}\left(âˆ’\mathrm{m}+\mathrm{n}\right)\\ â‡’\text{}2\mathrm{a}=\mathrm{d}\\ \mathrm{Now},\\ \frac{{\mathrm{T}}_{\mathrm{m}}}{{\mathrm{T}}_{\mathrm{n}}}=\frac{\mathrm{a}+\left(\mathrm{m}âˆ’1\right)2\mathrm{a}}{\mathrm{a}+\left(\mathrm{n}âˆ’1\right)2\mathrm{a}}\\ =\frac{\mathrm{a}+\left(\mathrm{m}âˆ’1\right)2\mathrm{a}}{\mathrm{a}+\left(\mathrm{n}âˆ’1\right)2\mathrm{a}}\\ =\frac{1+2\mathrm{m}âˆ’2}{1+2\mathrm{n}âˆ’2}\\ =\frac{2\mathrm{m}âˆ’1}{2\mathrm{n}âˆ’1}\\ \mathrm{Thus},{\text{the ratio of m}}^{\text{th}}{\text{and n}}^{\text{th}}\text{term is}\left(2\mathrm{m}âˆ’1\right):\left(2\mathrm{n}âˆ’1\right).\end{array}$

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

$\begin{array}{l}\mathrm{It}\text{is given that}\\ {\text{S}}_{\text{n}}={\text{3n}}^{\text{2}}+\text{5n}\\ \text{Replace n by n}âˆ’\text{1, we get}\\ {\text{S}}_{\text{n-1}}=\text{3}{\left(\text{n}âˆ’\text{1}\right)}^{\text{2}}+\text{5}\left(\text{n}âˆ’1\right)\\ \text{}={\text{3n}}^{\text{2}}âˆ’6\text{n}+\text{3}+\text{5n}âˆ’5\\ \text{}={\text{3n}}^{\text{2}}âˆ’\text{n}âˆ’2\\ \mathrm{So},{\text{T}}_{\text{n}}={\text{S}}_{\text{n}}âˆ’{\text{S}}_{\text{n-1}}\\ \text{}=\left({\text{3n}}^{\text{2}}+\text{5n}\right)âˆ’\left({\text{3n}}^{\text{2}}âˆ’\text{n}âˆ’2\right)\\ \text{}=6\text{n}+2\\ \mathrm{Since},{\text{T}}_{\text{m}}=164\\ \mathrm{So},\text{6m}+\text{2}=\text{164}\\ â‡’\text{m}=\frac{164âˆ’2}{6}\\ \text{}=\frac{162}{6}\\ \text{}=27\\ \mathrm{Thus},\text{the value of m is 27.}\end{array}$

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

Ans

$\begin{array}{l}\mathrm{Let}{\text{5 numbers between 8 and 26 are A}}_{\text{1}}{\text{, A}}_{\text{2}}{\text{, A}}_{\text{3}}{\text{, A}}_{\text{4}}{\text{and A}}_{\text{5}}\text{.}\\ {\text{Then, A.P. is 8, A}}_{\text{1}}{\text{, A}}_{\text{2}}{\text{, A}}_{\text{3}}{\text{, A}}_{\text{4}}{\text{, A}}_{\text{5}}\text{,â€‰26.}\\ Ã¢Ë†Âµ\mathrm{Last}\text{term}=\mathrm{a}+\left(\mathrm{n}âˆ’1\right)\mathrm{d}\\ \text{â€‰â€‰â€‰â€‰â€‰â€‰}26=8+\left(7âˆ’1\right)\mathrm{d}\end{array}$ $\begin{array}{l}\text{â€‰ â€‰â€‰â€‰}26=8+6\text{â€‰}\mathrm{d}\\ \text{â€‰ â€‰ â€‰ â€‰â€‰â€‰â€‰}\mathrm{d}=\frac{26âˆ’8}{6}\\ \text{â€‰â€‰â€‰}=\frac{18}{6}\\ \text{â€‰â€‰â€‰}=3\\ \mathrm{Then},\text{â€‰}{\mathrm{A}}_{1}=8+\mathrm{d}\\ \text{â€‰â€‰â€‰}=8+3\\ \text{â€‰â€‰â€‰}=11\\ \text{â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰}{\mathrm{A}}_{2}=8+2\mathrm{d}\\ \text{â€‰â€‰â€‰}=8+2\left(3\right)\\ \text{â€‰â€‰â€‰}=14\\ \text{â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰}{\mathrm{A}}_{3}=8+3\mathrm{d}\\ \text{â€‰â€‰â€‰}=8+3\left(3\right)\\ \text{â€‰â€‰â€‰}=17\\ \text{â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰}{\mathrm{A}}_{4}=8+4\mathrm{d}\\ \text{â€‰â€‰â€‰}=8+4\left(3\right)\\ \text{â€‰â€‰â€‰}=20\end{array}$ $\begin{array}{l}\end{array}$ $\begin{array}{l}\text{â€‰â€‰}{\mathrm{A}}_{5}=8+5\mathrm{d}\\ \text{â€‰â€‰â€‰}=8+5\left(3\right)\\ \text{â€‰â€‰â€‰}=23\\ \mathrm{Thus},\text{the 5 terms between 8 and 26 are 11, 14, 17, 20 and 23.}\end{array}$

Q.24

$\mathbf{\text{If}}\frac{{\mathbf{\text{a}}}^{\mathbf{\text{n}}}{\mathbf{\text{+b}}}^{\mathbf{\text{n}}}}{{\mathbf{\text{a}}}^{\mathbf{\text{n}}\mathbf{âˆ’}\mathbf{\text{1}}}{\mathbf{\text{+b}}}^{\mathbf{\text{n}}\mathbf{âˆ’}\mathbf{\text{1}}}}\mathbf{\text{â€‰â€‰is the A.M. between a and b, then find the value of n.}}$

Ans

$\begin{array}{l}\mathrm{A}.\mathrm{M}.\text{of a and b}=\frac{\mathrm{a}+\mathrm{b}}{2}\\ \text{â€‰}\frac{{\mathrm{a}}^{\mathrm{n}}+{\mathrm{b}}^{\mathrm{n}}}{{\mathrm{a}}^{\mathrm{n}–1}+{\mathrm{b}}^{\mathrm{n}–1}}=\frac{\mathrm{a}+\mathrm{b}}{2}\\ â‡’\left(\mathrm{a}+\mathrm{b}\right)\left({\mathrm{a}}^{\mathrm{n}–1}+{\mathrm{b}}^{\mathrm{n}–1}\right)\text{â€‰}=2\left({\mathrm{a}}^{\mathrm{n}}+{\mathrm{b}}^{\mathrm{n}}\right)\\ â‡’{\mathrm{a}}^{\mathrm{n}}+{\mathrm{ab}}^{\mathrm{n}âˆ’1}+{\mathrm{ba}}^{\mathrm{n}âˆ’1}+{\mathrm{b}}^{\mathrm{n}}=2{\mathrm{a}}^{\mathrm{n}}+2{\mathrm{b}}^{\mathrm{n}}\\ â‡’\text{â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰}{\mathrm{ab}}^{\mathrm{n}âˆ’1}âˆ’{\mathrm{b}}^{\mathrm{n}}={\mathrm{a}}^{\mathrm{n}}âˆ’{\mathrm{ba}}^{\mathrm{n}âˆ’1}\\ â‡’\text{â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰}{\mathrm{b}}^{\mathrm{n}âˆ’1}\left(\mathrm{a}âˆ’\mathrm{b}\right)=\text{â€‰}{\mathrm{a}}^{\mathrm{n}âˆ’1}\left(\mathrm{a}âˆ’\mathrm{b}\right)\\ â‡’\text{â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰}{\mathrm{b}}^{\mathrm{n}âˆ’1}=\text{â€‰}{\mathrm{a}}^{\mathrm{n}âˆ’1}\\ â‡’\text{â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰}{\left(\frac{\mathrm{a}}{\mathrm{b}}\right)}^{\mathrm{n}âˆ’1}=1=\text{â€‰}{\left(\frac{\mathrm{a}}{\mathrm{b}}\right)}^{0}\\ â‡’\text{â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰}\mathrm{n}=1\end{array}$

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 â€“ 1)th numbers is 5 : 9. Find the value of m.

Ans

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

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

Ans

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

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

$The\text{given G}\text{.P}\text{. is:}\text{}\text{}\text{}\frac{5}{2},\frac{5}{4},\frac{5}{8},\dots$ $\begin{array}{l}\mathrm{a}=\frac{5}{2}\\ \mathrm{r}=\frac{\left(\frac{5}{4}\right)}{\left(\frac{5}{2}\right)}=\frac{1}{2}\\ {\mathrm{T}}_{20}={\mathrm{ar}}^{20âˆ’1}\left[Ã¢Ë†Âµ{\mathrm{T}}_{\mathrm{n}}={\mathrm{ar}}^{\mathrm{n}âˆ’1}\right]\\ =\frac{5}{2}{\left(\frac{1}{2}\right)}^{19}\\ =\frac{5}{{2}^{20}}\\ \mathrm{and}\text{â€‰}{\mathrm{T}}_{\mathrm{n}}={\mathrm{ar}}^{\mathrm{n}âˆ’1}\\ =\frac{5}{2}{\left(\frac{1}{2}\right)}^{\mathrm{n}âˆ’1}\\ =\frac{5}{{2}^{\mathrm{n}}}\end{array}$

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

Ans

$\begin{array}{l}\mathrm{Common}\text{ratio of given G.P.}=2\\ {8}^{\mathrm{th}}\text{term of A.P.}=192\\ \text{â€‰}{\mathrm{ar}}^{8âˆ’1}=192\\ \text{â€‰â€‰â€‰}\mathrm{a}.{2}^{7}=192\\ \text{â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰}\mathrm{a}=\frac{192}{{2}^{7}}=1.5\\ {12}^{\mathrm{th}}\text{term of A.P.}={\mathrm{ar}}^{12âˆ’1}\end{array}$ $\begin{array}{l}=1.5{\left(2\right)}^{11}\\ =1.5Ã—2048\\ =3072\end{array}$

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

Ans

$\begin{array}{l}\text{Let first term of G.P. be a and common ratio be r.}\\ {\text{â€‰â€‰5}}^{\text{th}}\text{term of G.P.}=\mathrm{p}\\ â‡’\text{â€‰â€‰}{\mathrm{T}}_{5}=\mathrm{p}\\ â‡’\text{â€‰â€‰â€‰â€‰}\mathrm{p}={\mathrm{ar}}^{4}\\ {\text{â€‰â€‰8}}^{\text{th}}\text{term of G.P.}=\mathrm{q}\\ â‡’\text{â€‰â€‰}{\mathrm{T}}_{8}=\mathrm{q}\\ â‡’\text{â€‰â€‰â€‰â€‰}\mathrm{q}={\mathrm{ar}}^{7}\\ {\text{11}}^{\text{th}}\text{term of G.P.}=\mathrm{s}\\ â‡’{\mathrm{T}}_{11}=\mathrm{s}\\ â‡’\text{â€‰ â€‰â€‰}\mathrm{s}={\mathrm{ar}}^{10}\\ \text{L.H.S.}={\mathrm{q}}^{2}\\ \text{â€‰ â€‰â€‰â€‰â€‰â€‰}={\left({\mathrm{ar}}^{7}\right)}^{2}\\ \text{â€‰ â€‰â€‰â€‰â€‰â€‰}={\mathrm{a}}^{2}{\mathrm{r}}^{14}\\ \text{R.H.S.=}\mathrm{ps}\\ \text{â€‰ â€‰â€‰â€‰â€‰â€‰}=\left({\mathrm{ar}}^{4}\right)\left({\mathrm{ar}}^{10}\right)\\ \text{â€‰ â€‰â€‰â€‰â€‰â€‰}={\mathrm{a}}^{2}{\mathrm{r}}^{14}\\ \text{So,â€‰â€‰â€‰â€‰â€‰L.H.S.=R.H.S.}\\ \text{Hence proved.}\end{array}$

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

Ans

$\begin{array}{l}\text{â€‰â€‰â€‰â€‰â€‰â€‰â€‰}\mathrm{First}\text{term of G.P.}=âˆ’3\\ \mathrm{Let}\text{common ratio of G.P.}=\mathrm{r}\\ \mathrm{According}\text{to given condition,}\\ {\mathrm{T}}_{4}={\left({\mathrm{T}}_{2}\right)}^{2}\\ â‡’\text{â€‰}{\mathrm{ar}}^{3}={\left(\mathrm{ar}\right)}^{2}\\ â‡’\text{â€‰â€‰}\left(âˆ’3\right){\mathrm{r}}^{3}={\left(âˆ’3\right)}^{2}{\mathrm{r}}^{2}\\ â‡’\text{â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰}\mathrm{r}=âˆ’3\\ \mathrm{Then},{\mathrm{T}}_{7}={\mathrm{ar}}^{6}\\ \text{â€‰â€‰â€‰â€‰â€‰â€‰}=\left(âˆ’3\right){\left(âˆ’3\right)}^{6}\\ \text{â€‰â€‰â€‰â€‰â€‰â€‰}={\left(âˆ’3\right)}^{7}\\ \text{â€‰â€‰â€‰â€‰â€‰â€‰}=âˆ’2187\\ \mathrm{Thus},{\text{7}}^{\text{th}}\text{term of G.P. is}âˆ’2187.\end{array}$

Q.31

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

Ans

$\begin{array}{l}\left(\mathrm{a}\right)\mathrm{Let}{\text{t}}_{\text{n}}=128,\text{a}=2,\\ \text{â€‰â€‰â€‰ r}=\frac{2\sqrt{2}}{2}=\sqrt{2}\end{array}$ $\begin{array}{l}\text{â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰}{\mathrm{t}}_{\mathrm{n}}={\mathrm{ar}}^{\mathrm{n}âˆ’1}\\ \text{â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰}128=2{\left(\sqrt{2}\right)}^{\mathrm{n}âˆ’1}\\ â‡’\text{â€‰â€‰â€‰â€‰â€‰â€‰}{2}^{7}={\left(2\right)}^{1+\frac{\mathrm{n}âˆ’1}{2}}\\ â‡’\text{â€‰ â€‰â€‰â€‰â€‰â€‰â€‰â€‰}7=1+\frac{\mathrm{n}âˆ’1}{2}\\ â‡’\text{â€‰}6Ã—2=\mathrm{n}âˆ’1â‡’\text{â€‰â€‰â€‰â€‰â€‰â€‰â€‰}\mathrm{n}=13\\ \mathrm{Thus},{\text{128 is 13}}^{\text{th}}\text{term of G.P.}\\ \left(\mathrm{b}\right)\text{â€‰â€‰}\mathrm{Let}{\text{t}}_{\text{n}}=729,\text{a}=\sqrt{3},\\ \text{â€‰â€‰â€‰ r}=\frac{3}{\sqrt{3}}=\sqrt{3}\\ \text{â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰}{\mathrm{t}}_{\mathrm{n}}={\mathrm{ar}}^{\mathrm{n}âˆ’1}\\ \text{â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰}729=\sqrt{3}{\left(\sqrt{3}\right)}^{\mathrm{n}âˆ’1}\\ â‡’\text{â€‰â€‰â€‰â€‰â€‰â€‰â€‰}{3}^{6}={\left(3\right)}^{\frac{1}{2}+\frac{\mathrm{n}âˆ’1}{2}}\\ â‡’\text{â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰}6=\frac{1}{2}+\frac{\mathrm{n}âˆ’1}{2}\\ â‡’\text{â€‰}6Ã—2=1+\mathrm{n}âˆ’1\\ â‡’\text{â€‰â€‰â€‰â€‰â€‰â€‰â€‰}\mathrm{n}=12\\ \mathrm{Thus},\text{}729{\text{is 12}}^{\text{th}}\text{term of G.P.}\\ \left(\mathrm{c}\right)\text{â€‰}\mathrm{Let}{\text{t}}_{\text{n}}=729,\text{a}=\frac{1}{3},\\ \text{â€‰â€‰â€‰r}=\frac{\left(\frac{1}{9}\right)}{\left(\frac{1}{3}\right)}=\frac{1}{3}\end{array}$ $\begin{array}{l}\text{â€‰}\text{â€‰}\text{â€‰}\text{â€‰}\text{â€‰}\text{â€‰}\text{â€‰}\text{â€‰}\text{â€‰}\text{â€‰}\text{â€‰}\text{â€‰}\end{array}$ $\begin{array}{l}{\mathrm{t}}_{\mathrm{n}}={\mathrm{ar}}^{\mathrm{n}âˆ’1}\\ \text{â€‰â€‰â€‰}\frac{1}{19683}=\frac{1}{3}{\left(\frac{1}{3}\right)}^{\mathrm{n}âˆ’1}\\ â‡’\text{â€‰â€‰â€‰â€‰â€‰â€‰â€‰}{\left(\frac{1}{3}\right)}^{9}={\left(\frac{1}{3}\right)}^{\mathrm{n}}\\ â‡’\text{â€‰â€‰â€‰â€‰â€‰â€‰â€‰ â€‰}9=\mathrm{n}\\ â‡’\text{â€‰â€‰â€‰â€‰â€‰â€‰â€‰}\mathrm{n}=9\\ \mathrm{Thus},\text{}\frac{1}{19683}{\text{is 9}}^{\text{th}}\text{term of G.P.}\end{array}$

Q.32

$\mathbf{\text{For what values of x, the numbersâ€‰â€‰}}\mathbf{âˆ’}\frac{\mathbf{\text{2}}}{\mathbf{\text{7}}}\mathbf{\text{,x,}}\mathbf{âˆ’}\frac{\mathbf{\text{7}}}{\mathbf{\text{2}}}\mathbf{\text{are in G.P.?}}$

Ans

$\begin{array}{l}âˆ’\frac{\text{2}}{\text{7}}\text{,x,}âˆ’\frac{\text{7}}{\text{2}}\text{are in G.P., then}\\ \text{Common ratio}=\frac{\mathrm{x}}{âˆ’\left(\frac{2}{7}\right)}\\ \text{â€‰â€‰}=âˆ’\frac{7\mathrm{x}}{2}\\ \text{Also, common ratio}=\frac{\left(âˆ’\frac{7}{2}\right)}{\mathrm{x}}\\ \text{â€‰â€‰}=âˆ’\frac{7}{2\mathrm{x}}\\ \text{Then,}\\ âˆ’\frac{7\mathrm{x}}{2}=âˆ’\frac{7}{2\mathrm{x}}\\ â‡’\text{â€‰â€‰â€‰â€‰}{\mathrm{x}}^{2}=1\\ â‡’\text{â€‰â€‰â€‰â€‰â€‰}\mathrm{x}=Â±1\end{array}$

Q.33

Ans

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

Q.34

$\mathbf{Evaluate}\mathbf{:}\mathbf{\text{â€‰â€‰}}\underset{k=1}{\overset{11}{âˆ‘}}\left(2+{3}^{k}\right).$

Ans

$\begin{array}{l}\mathrm{We}\mathrm{}\mathrm{have},\\ \underset{\mathrm{k}=1}{\overset{11}{âˆ‘}}\left(2+{3}^{\mathrm{k}}\right)=\left(2+{3}^{1}\right)+\left(2+{3}^{2}\right)+\left(2+{3}^{3}\right)+...+\left(2+{3}^{11}\right)\\ \text{â€‰}=2Ã—11+\left({3}^{1}+{3}^{2}+{3}^{3}+...+{3}^{11}\right)\\ \text{â€‰}=22+\frac{3\left({3}^{11}âˆ’1\right)}{3âˆ’1}\\ \text{â€‰}=22+\frac{3}{2}\left({3}^{11}âˆ’1\right)\end{array}$

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

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

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

Ans

$\begin{array}{l}\text{G}.\text{P}.\text{is 3},{\text{3}}^{\text{2}},{\text{3}}^{\text{3}},\text{}â€¦\\ \mathrm{a}=3,\text{r}=\frac{{3}^{2}}{3}=3{\text{and S}}_{\text{n}}=120\\ {\mathrm{S}}_{\mathrm{n}}=\mathrm{a}\frac{\left({\mathrm{r}}^{\mathrm{n}}âˆ’1\right)}{\left(\mathrm{r}âˆ’1\right)}\end{array}$ $\begin{array}{l}120=3\left(\frac{{3}^{\mathrm{n}}âˆ’1}{3âˆ’1}\right)\\ 120=\frac{3}{2}\left({3}^{\mathrm{n}}âˆ’1\right)\\ 120Ã—\frac{2}{3}=\left({3}^{\mathrm{n}}âˆ’1\right)\\ \text{â€‰}81={3}^{\mathrm{n}}\\ â‡’\text{â€‰â€‰â€‰â€‰â€‰â€‰}{3}^{4}={3}^{\mathrm{n}}\\ â‡’\text{â€‰â€‰â€‰}\mathrm{n}=4\\ \mathrm{Thus},\text{4 terms of G.P. are needed to give the sum 120.}\end{array}$

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

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

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

Ans

$\begin{array}{l}\mathrm{First}\text{term of G.P.}=\left(\text{a}\right)\\ \text{â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰}=729\\ {7}^{\mathrm{th}}\text{}\mathrm{term}\text{of G.P.}={\mathrm{T}}_{7}\\ \text{â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰}=64\\ \mathrm{Then},{\text{â€‰ar}}^{\text{6}}=64\\ â‡’\text{â€‰â€‰}729\text{â€‰}{\mathrm{r}}^{6}=64\\ \text{â€‰â€‰â€‰â€‰}{\mathrm{r}}^{6}=\frac{64}{729}\\ \text{â€‰â€‰â€‰â€‰}{\mathrm{r}}^{6}=\frac{{2}^{6}}{{3}^{6}}\\ â‡’\text{â€‰â€‰â€‰â€‰â€‰â€‰}\mathrm{r}=\frac{2}{3}\\ {\mathrm{S}}_{\mathrm{n}}=\frac{\mathrm{a}\left(1âˆ’{\mathrm{r}}^{\mathrm{n}}\right)}{\left(1âˆ’\mathrm{r}\right)}\\ â‡’\text{â€‰}{\mathrm{S}}_{7}=\frac{729\left(1âˆ’\frac{{2}^{\mathrm{n}}}{{3}^{\mathrm{n}}}\right)}{\left(1âˆ’\frac{2}{3}\right)}\end{array}$ $\begin{array}{l}â‡’\text{â€‰â€‰}{\mathrm{S}}_{7}=729Ã—3\left(1âˆ’\frac{{2}^{7}}{{3}^{7}}\right)\\ â‡’\text{â€‰â€‰}{\mathrm{S}}_{7}=729Ã—3\left(1âˆ’\frac{128}{2187}\right)\\ â‡’\text{â€‰â€‰}{\mathrm{S}}_{7}=2187\left(\frac{2187âˆ’128}{2187}\right)\\ â‡’\text{â€‰â€‰}{\mathrm{S}}_{7}=2059\\ \mathrm{Thus},\text{sum of 7 terms of G.P. is 2059.}\end{array}$

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

Ans

$\begin{array}{l}\mathrm{Let}\text{G.P. be:}\\ {\text{a, ar, ar}}^{\text{2}}{\text{, ar}}^{\text{3}}{\text{, ar}}^{\text{4}}{\text{, ar}}^{\text{5}}\text{,}...\\ \text{Then according to first condition,}\\ \text{â€‰â€‰â€‰ a}+\mathrm{ar}=âˆ’\text{â€‰}4\\ \mathrm{or}\text{â€‰}\mathrm{a}\left(1+\mathrm{r}\right)=âˆ’4\\ \mathrm{or}\text{â€‰â€‰â€‰}\mathrm{a}=\frac{âˆ’4}{1+\mathrm{r}}...\left(\mathrm{i}\right)\\ \text{Then according to second condition,}\\ {\text{T}}_{\text{5}}={\text{4T}}_{\text{3}}\\ â‡’{\text{â€‰ar}}^{\text{4}}=4\left({\text{ar}}^{\text{2}}\right)\\ â‡’\text{â€‰â€‰}{\mathrm{r}}^{2}=4\\ â‡’\text{â€‰â€‰â€‰}\mathrm{r}=Â±\text{â€‰}2\\ \mathrm{Putting}\text{r}=2\text{in equation}\left(\mathrm{i}\right),\text{we get}\end{array}$ $\begin{array}{l}\text{â€‰â€‰â€‰}\mathrm{a}=\frac{âˆ’4}{1+2}\\ \text{â€‰â€‰â€‰â€‰â€‰â€‰}=âˆ’\frac{4}{3}\\ \mathrm{Putting}\text{r}=2\text{in equation}\left(\mathrm{i}\right),\text{we get}\\ \text{â€‰â€‰â€‰}\mathrm{a}=\frac{âˆ’4}{1âˆ’2}\\ \text{â€‰â€‰â€‰â€‰â€‰â€‰}=\frac{âˆ’4}{âˆ’1}\\ \text{â€‰â€‰â€‰â€‰â€‰â€‰}=4\\ \mathrm{Therefore},\text{G.P. when a}=âˆ’\frac{4}{3}\text{and r}=\text{2}\\ âˆ’\frac{4}{3}\text{,}âˆ’\frac{4}{3}\left(2\right)\text{,}âˆ’\frac{4}{3}{\left(2\right)}^{\text{2}}\text{,}âˆ’\frac{4}{3}{\left(2\right)}^{\text{3}}\text{,}âˆ’\frac{4}{3}{\left(2\right)}^{\text{4}}\text{,}âˆ’\frac{4}{3}{\left(2\right)}^{\text{5}}\text{,}...\\ \text{orâ€‰â€‰â€‰â€‰â€‰â€‰â€‰}âˆ’\frac{4}{3},âˆ’\frac{8}{3},âˆ’\frac{16}{3},âˆ’\frac{32}{3},âˆ’\frac{64}{3},âˆ’\frac{128}{3},...\\ \mathrm{Therefore},\text{G.P. when a}=4\text{and r}=âˆ’\text{2}\\ 4\text{,}4\left(âˆ’2\right)\text{,}4{\left(âˆ’2\right)}^{\text{2}}\text{,}4{\left(âˆ’2\right)}^{\text{3}}\text{,}4{\left(âˆ’2\right)}^{\text{4}}\text{,}4{\left(âˆ’2\right)}^{\text{5}}\text{,}...\\ \text{orâ€‰â€‰}4,âˆ’8,16,âˆ’32,...\\ \mathrm{Therefore},\text{the required G.P. is}\\ \text{â€‰}âˆ’\frac{4}{3},âˆ’\frac{8}{3},âˆ’\frac{16}{3},âˆ’\frac{32}{3},âˆ’\frac{64}{3},âˆ’\frac{128}{3},...\\ \mathrm{or}\text{â€‰}4,âˆ’8,16,âˆ’32,...\end{array}$

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

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

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

Ans

$\begin{array}{l}\mathrm{Let}\text{â€‰}{\mathrm{S}}_{\mathrm{n}}=\text{8}+\text{88}+\text{888}+\text{8888}+â€¦\\ \text{â€‰â€‰â€‰â€‰}=8\left(1+11+111+1111+...\right)\\ \text{â€‰â€‰â€‰â€‰}=\frac{8}{9}\left(9+99+999+9999+...\right)\\ \text{â€‰â€‰â€‰â€‰}=\frac{8}{9}\left\{\left(10âˆ’1\right)+\left({10}^{2}âˆ’1\right)+\left({10}^{3}âˆ’1\right)+\left({10}^{4}âˆ’1\right)+...\right\}\\ \text{â€‰â€‰â€‰â€‰}=\frac{8}{9}\left\{10+{10}^{2}+{10}^{3}+{10}^{4}+...âˆ’1âˆ’1âˆ’1âˆ’1âˆ’...\text{â€‰}\mathrm{n}\text{times}\right\}\end{array}$ $\begin{array}{l}\text{â€‰â€‰â€‰â€‰}=\frac{8}{9}\left\{\frac{10\left({10}^{\mathrm{n}}âˆ’1\right)}{10âˆ’1}âˆ’\text{â€‰}\mathrm{n}\right\}\left[Ã¢Ë†Âµ{\mathrm{S}}_{\mathrm{n}}=\frac{\mathrm{a}\left({\mathrm{r}}^{\mathrm{n}}âˆ’1\right)}{\mathrm{r}âˆ’1}\right]\\ \text{â€‰â€‰â€‰â€‰}=\frac{8}{9}\left\{\frac{10\left({10}^{\mathrm{n}}âˆ’1\right)}{9}âˆ’\text{â€‰}\mathrm{n}\right\}\\ \text{â€‰â€‰â€‰â€‰}=\frac{80}{81}\left({10}^{\mathrm{n}}âˆ’1\right)âˆ’\frac{8}{9}\mathrm{n}\end{array}$

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

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

Q.43 Show that the products of the corresponding terms of the sequences a, ar, ar2, â€¦ arn â€“ 1 and A, AR, AR2, â€¦ ARnâ€“1 form a G.P, and find the common ratio.

Ans

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

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

$\begin{array}{l}{\text{Let four numbers of G.P. are a, ar, ar}}^{\text{2}}{\text{, ar}}^{\text{3}}\text{.}\\ \text{According to first condition,}\\ {\text{â€‰â€‰ T}}_{\text{3}}âˆ’{\text{T}}_{\text{1}}=9\\ â‡’\text{â€‰}{\mathrm{ar}}^{2}âˆ’\mathrm{a}=9\\ â‡’\text{â€‰}\mathrm{a}\left({\mathrm{r}}^{2}âˆ’1\right)=9...\left(\mathrm{i}\right)\\ \text{According to second condition,}\\ {\text{T}}_{\text{2}}âˆ’{\text{T}}_{\text{4}}=18\\ â‡’\text{â€‰}\mathrm{ar}âˆ’{\mathrm{ar}}^{3}=18\\ â‡’\text{â€‰â€‰â€‰}\mathrm{ar}\left(1âˆ’{\mathrm{r}}^{2}\right)=18\\ â‡’\text{â€‰â€‰â€‰}\mathrm{ar}\left({\mathrm{r}}^{2}âˆ’1\right)=âˆ’18...\left(\mathrm{ii}\right)\\ \text{Dividing equation}\left(\text{ii}\right)\text{by equationâ€‰}\left(\text{i}\right)\text{, we get}\\ \text{â€‰â€‰â€‰â€‰â€‰â€‰â€‰}\frac{\mathrm{ar}\left({\mathrm{r}}^{2}âˆ’1\right)}{\mathrm{a}\left({\mathrm{r}}^{2}âˆ’1\right)}=âˆ’\frac{18}{9}\\ â‡’\text{â€‰â€‰â€‰â€‰â€‰}\mathrm{r}=âˆ’2\\ \text{Putting r=}âˆ’\text{2, in equation}\left(\text{i}\right)\text{, we get}\\ \mathrm{a}\left\{{\left(âˆ’2\right)}^{2}âˆ’1\right\}=9\\ \text{â€‰â€‰â€‰â€‰â€‰}\mathrm{a}=\frac{9}{3}\\ \text{â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰}=3\\ \text{Thus, the four terms of G.P. are:}\\ \text{3, 3}\left(âˆ’2\right)\text{, 3}{\left(âˆ’2\right)}^{\text{2}}\text{, 3}{\left(âˆ’2\right)}^{\text{3}}\\ \mathrm{i}.\mathrm{e}.,\text{â€‰â€‰â€‰â€‰}3,âˆ’6,12,âˆ’24\end{array}$

Q.45

$\begin{array}{l}{\mathbf{\text{If the p}}}^{\mathbf{\text{th}}}{\mathbf{\text{, q}}}^{\mathbf{\text{th}}}{\mathbf{\text{â€‰and r}}}^{\mathbf{\text{th}}}\mathbf{\text{terms of a G.P. are a and b, respectively.}}\\ {\mathbf{\text{Prove thatâ€‰ a}}}^{\mathbf{\text{q}}\mathbf{âˆ’}\mathbf{\text{r}}}{\mathbf{\text{â€‰b}}}^{\mathbf{\text{r}}\mathbf{âˆ’}\mathbf{\text{p}}}{\mathbf{\text{â€‰c}}}^{\mathbf{\text{p}}\mathbf{âˆ’}\mathbf{\text{q}}}\mathbf{\text{=1.}}\end{array}$

Ans

$\begin{array}{l}\mathrm{Let}\text{first term of G.P. is A and common ratio is R.}\\ \mathrm{Given}\text{that:}\\ {\text{T}}_{\text{p}}=\text{a}â‡’{\mathrm{AR}}^{\mathrm{p}âˆ’1}=\mathrm{a}\\ {\text{T}}_{\text{q}}=\text{b}â‡’{\mathrm{AR}}^{\mathrm{q}âˆ’1}=\mathrm{b}\\ {\text{T}}_{\text{r}}=\text{c}â‡’{\mathrm{AR}}^{\mathrm{r}âˆ’1}=\mathrm{c}\\ \mathrm{L}.\mathrm{H}.\mathrm{S}.\text{â€‰}={\mathrm{a}}^{\mathrm{q}–\mathrm{r}}\text{â€‰}{\mathrm{b}}^{\mathrm{r}–\mathrm{p}}\text{â€‰}{\mathrm{c}}^{\mathrm{p}–\mathrm{q}}\\ \text{â€‰â€‰â€‰â€‰â€‰ â€‰}={\left({\mathrm{AR}}^{\mathrm{p}âˆ’1}\right)}^{\mathrm{q}–\mathrm{r}}\text{â€‰}{\left({\mathrm{AR}}^{\mathrm{q}âˆ’1}\right)}^{\mathrm{r}–\mathrm{p}}\text{â€‰}{\left({\mathrm{AR}}^{\mathrm{r}âˆ’1}\right)}^{\mathrm{p}–\mathrm{q}}\\ \text{â€‰â€‰â€‰â€‰â€‰â€‰}={\mathrm{A}}^{\mathrm{q}âˆ’\mathrm{r}+\mathrm{r}âˆ’\mathrm{p}+\mathrm{p}âˆ’\mathrm{q}}\text{â€‰}{\mathrm{R}}^{\left(\mathrm{p}âˆ’1\right)\left(\mathrm{q}âˆ’\mathrm{r}\right)+\left(\mathrm{q}âˆ’1\right)\left(\mathrm{r}âˆ’\mathrm{p}\right)\left(\mathrm{r}âˆ’1\right)\left(\mathrm{p}âˆ’\mathrm{q}\right)}\\ \text{â€‰â€‰â€‰â€‰â€‰â€‰}={\mathrm{A}}^{0}{\mathrm{R}}^{\mathrm{pq}âˆ’\mathrm{pr}âˆ’\mathrm{q}+\mathrm{r}+\mathrm{qr}âˆ’\mathrm{qp}âˆ’\mathrm{r}+\mathrm{p}+\mathrm{rp}âˆ’\mathrm{rq}âˆ’\mathrm{p}+\mathrm{q}}\\ \text{â€‰â€‰â€‰â€‰â€‰â€‰}={\mathrm{A}}^{0}{\mathrm{R}}^{0}\\ \text{â€‰â€‰â€‰â€‰â€‰â€‰}=1Ã—1\\ \text{â€‰â€‰â€‰â€‰â€‰â€‰}=1=\mathrm{R}.\mathrm{H}.\mathrm{S}.\end{array}$

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

$\begin{array}{l}\mathrm{First}\text{term of G.P.}=\text{a}\\ {\text{â€‰â€‰â€‰â€‰n}}^{\text{th}}\text{term of G.P.}=\text{b}\\ \text{Let common ratio}=\mathrm{r}\\ \mathrm{The}\text{G.P. is}\\ {\text{a, ar, ar}}^{\text{2}}{\text{, ar}}^{\text{3}}{\text{, ar}}^{\text{4}}\text{,}...\text{, b}\\ \text{and b}={\text{ar}}^{\text{n-1}}\\ \text{Product of n terms}=\text{a}Ã—\text{ar}Ã—{\text{ar}}^{\text{2}}Ã—{\text{ar}}^{\text{3}}Ã—{\text{ar}}^{\text{4}}Ã—...Ã—\text{b}\\ \text{â€‰â€‰}=\text{a}Ã—\text{ar}Ã—{\text{ar}}^{\text{2}}Ã—{\text{ar}}^{\text{3}}Ã—{\text{ar}}^{\text{4}}Ã—...Ã—{\text{ar}}^{\text{n-1}}\end{array}$ $\begin{array}{l}\text{â€‰â€‰}={\mathrm{a}}^{\mathrm{n}}Ã—{\mathrm{r}}^{1+2+3+4+...+\left(\mathrm{n}âˆ’1\right)}\\ \text{â€‰â€‰â€‰â€‰â€‰â€‰â€‰}\mathrm{P}\text{â€‰}={\mathrm{a}}^{\mathrm{n}}Ã—{\mathrm{r}}^{\frac{\left(\mathrm{n}âˆ’1\right)\left(\mathrm{n}âˆ’1+1\right)}{2}}\\ \text{â€‰â€‰â€‰â€‰â€‰â€‰â€‰}\mathrm{P}\text{â€‰}={\mathrm{a}}^{\mathrm{n}}Ã—{\mathrm{r}}^{\frac{\mathrm{n}\left(\mathrm{n}âˆ’1\right)}{2}}\\ \mathrm{Squarring}\text{both sides, we get}\\ {\text{P}}^{\text{2}}={\mathrm{a}}^{2\mathrm{n}}{\mathrm{r}}^{\mathrm{n}\left(\mathrm{n}âˆ’1\right)}\\ ={\left({\mathrm{a}}^{2}{\mathrm{r}}^{\mathrm{n}âˆ’1}\right)}^{\mathrm{n}}\\ ={\left(\mathrm{a}Ã—{\mathrm{ar}}^{\mathrm{n}âˆ’1}\right)}^{\mathrm{n}}\\ {\text{â€‰P}}^{\text{2}}={\left(\mathrm{ab}\right)}^{\mathrm{n}}\left[Ã¢Ë†Âµ\mathrm{b}={\mathrm{ar}}^{\mathrm{n}âˆ’1}\right]\\ \mathrm{Hence}\text{Proved.}\end{array}$

Q.47

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

Ans

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

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

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

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

Ans

$\begin{array}{l}\mathrm{Let}\text{two numbers between 3 and 81 are x and y in such a way}\\ \text{that 3, x, y, 81 are in G.P.}\\ \text{So, first term of G.P.}=\text{3}\\ \text{â€‰}\mathrm{Common}\text{difference}=\mathrm{r}\\ \text{â€‰â€‰}{4}^{\mathrm{th}}\text{â€‰}\mathrm{term}\text{of G.P.}={\mathrm{ar}}^{3}\\ â‡’\text{â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰}3\text{â€‰}{\mathrm{r}}^{3}=81\\ â‡’\text{â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰ â€‰â€‰}{\mathrm{r}}^{3}=\frac{81}{3}\\ â‡’\text{â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰}\mathrm{r}=\sqrt[3]{27}\\ \text{â€‰â€‰â€‰â€‰}=3\end{array}$ $\begin{array}{l}âˆ´\mathrm{x}={\mathrm{T}}_{2}\\ \text{â€‰â€‰â€‰ â€‰}=\mathrm{ar}\\ \text{â€‰â€‰â€‰â€‰}=3Ã—3\\ \text{â€‰â€‰â€‰ â€‰}=9\\ \mathrm{andy}={\mathrm{T}}_{3}\\ \text{â€‰â€‰â€‰â€‰}={\mathrm{ar}}^{2}\\ \text{â€‰â€‰â€‰â€‰}=3Ã—{3}^{2}\\ \text{â€‰â€‰â€‰â€‰}=27\\ \mathrm{Thus},\text{two numbers between 3 and 81 are 9 and 27.}\end{array}$

Q.50

$\mathbf{\text{Find the value of n so thatâ€‰}}\frac{{\mathbf{\text{a}}}^{\mathbf{\text{n+1}}}{\mathbf{\text{+b}}}^{\mathbf{\text{n+1}}}}{{\mathbf{\text{a}}}^{\mathbf{\text{n}}}{\mathbf{\text{+b}}}^{\mathbf{\text{n}}}}\mathbf{\text{may be the geometric mean between a and b.}}$

Ans

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

Q.51

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

Ans

$\begin{array}{l}\mathrm{Let}\text{two numbers be a and b.}\\ \text{G.M. of a and b}=\sqrt{\mathrm{ab}}\\ \mathrm{Then},\text{according to given condition,}\\ \text{â€‰â€‰â€‰ â€‰a}+\text{b}=6\sqrt{\mathrm{ab}}...\left(\mathrm{i}\right)\\ Ã¢Ë†Âµ\text{â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰}{\left(\text{a}âˆ’\text{b}\right)}^{2}={\left(\text{a}+\text{b}\right)}^{2}âˆ’4\mathrm{ab}\\ â‡’\text{â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰}{\left(\text{a}âˆ’\text{b}\right)}^{2}={\left(6\sqrt{\mathrm{ab}}\right)}^{2}âˆ’4\mathrm{ab}\\ \text{â€‰â€‰â€‰â€‰â€‰}=36\mathrm{ab}âˆ’4\mathrm{ab}\\ \text{â€‰â€‰â€‰â€‰â€‰}=32\mathrm{ab}\\ â‡’\text{â€‰â€‰â€‰â€‰ a}âˆ’\text{b}=\sqrt{32\mathrm{ab}}\\ â‡’\text{â€‰â€‰â€‰â€‰ a}âˆ’\text{b}=4\sqrt{2}\sqrt{\mathrm{ab}}...\left(\mathrm{ii}\right)\end{array}$ $\begin{array}{l}\mathrm{Solving}\text{equation}\left(\mathrm{i}\right)\text{and equation}\left(\mathrm{ii}\right),\text{we get}\\ \text{a}=\left(3+2\sqrt{2}\right)\sqrt{\mathrm{ab}}\text{and b}=\left(3âˆ’2\sqrt{2}\right)\sqrt{\mathrm{ab}}\\ \mathrm{Then},\\ \text{â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰ratio of the numbers}=\frac{\left(3+2\sqrt{2}\right)\sqrt{\mathrm{ab}}}{\left(3âˆ’2\sqrt{2}\right)\sqrt{\mathrm{ab}}}\\ \text{â€‰â€‰â€‰â€‰â€‰}=\left(3+2\sqrt{2}\right):\left(3âˆ’2\sqrt{2}\right)\end{array}$

Q.52

$\begin{array}{l}\mathbf{\text{If A and G be A.M. and G.M., respectively between two positive numbers,}}\\ \mathbf{\text{prove that the numbers areâ€‰â€‰AÂ±}}\sqrt{\left(\text{A+G}\right)\left(\text{A}âˆ’\text{G}\right)}\mathbf{\text{.}}\end{array}$

Ans

$\begin{array}{l}\text{Let two numbers are a and b.}\\ \text{Then, A.M.}=\frac{\mathrm{a}+\mathrm{b}}{2}\\ â‡’\mathrm{A}=\frac{\mathrm{a}+\mathrm{b}}{2}\\ â‡’\text{â€‰â€‰â€‰â€‰}\mathrm{a}+\mathrm{b}=2\mathrm{A}...\left(\mathrm{i}\right)\\ Ã¢Ë†Âµ\text{â€‰â€‰â€‰â€‰â€‰}\mathrm{G}.\mathrm{M}.=\sqrt{\mathrm{ab}}\\ â‡’\text{â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰}\mathrm{G}=\sqrt{\mathrm{ab}}\\ â‡’\text{â€‰â€‰â€‰â€‰â€‰â€‰â€‰ â€‰}{\mathrm{G}}^{2}=\mathrm{ab}...\left(\mathrm{ii}\right)\\ Ã¢Ë†Âµ\text{â€‰â€‰â€‰â€‰}\mathrm{a}âˆ’\mathrm{b}=\sqrt{{\left(\mathrm{a}+\mathrm{b}\right)}^{2}âˆ’4\mathrm{ab}}\\ \text{â€‰â€‰â€‰â€‰â€‰}=\sqrt{{\left(2\mathrm{A}\right)}^{2}âˆ’4{\mathrm{G}}^{2}}\\ \text{â€‰â€‰â€‰â€‰}\mathrm{a}âˆ’\mathrm{b}=2\sqrt{{\mathrm{A}}^{2}âˆ’{\mathrm{G}}^{2}}...\left(\mathrm{iii}\right)\\ \text{Solving equation}\left(\text{i}\right)\text{and equation}\left(\text{iii}\right)\text{, we get}\\ \text{a}=\mathrm{A}+\sqrt{{\mathrm{A}}^{2}âˆ’{\mathrm{G}}^{2}}\text{and b}=\mathrm{A}âˆ’\sqrt{{\mathrm{A}}^{2}âˆ’{\mathrm{G}}^{2}}\\ \text{Therefore, the two numbers are:}\\ \mathrm{A}Â±\sqrt{\left(\mathrm{A}+\mathrm{G}\right)\left(\mathrm{A}âˆ’\mathrm{G}\right)}.\end{array}$

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

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

Q.54 What will Ã¢â€šÂ¹ 500 amounts to in 10 years after its deposit in a bank which pays annual interest rate of 10% compounded annually?

Ans

$\begin{array}{l}\text{Principal amount deposited in the bank}=\text{â€‰}Ã¢â€šÂ¹\text{â€‰}500\\ \mathrm{Rate}\text{of interest in the bank}=10\mathrm{%}\text{â€‰}\mathrm{p}.\mathrm{a}.\\ \mathrm{Number}\text{of years for amount deposited}=10\text{years}\\ \mathrm{Amount}\text{after one year in the bank}=500\left(1+\frac{10}{100}\right)\\ =500\left(\frac{11}{10}\right)\\ \mathrm{Amount}\text{after two years in the bank}=500{\left(1+\frac{10}{100}\right)}^{2}\\ =500{\left(\frac{11}{10}\right)}^{2}\\ \mathrm{Amount}\text{after three years in the bank}=500{\left(1+\frac{10}{100}\right)}^{3}\\ =500{\left(\frac{11}{10}\right)}^{3}\\ \mathrm{So},\text{G.P. formed by using amount obtained from bank is}\end{array}$ $\begin{array}{l}500\left(\frac{11}{10}\right),500{\left(\frac{11}{10}\right)}^{2},500{\left(\frac{11}{10}\right)}^{3},...\\ \mathrm{First}\text{term of G.P.}=500\left(\frac{11}{10}\right)\\ \mathrm{Common}\text{ratio of G.P.}=\frac{11}{10}\\ \mathrm{Number}\text{of years}=10\\ \mathrm{Since},{\text{T}}_{\mathrm{n}}={\mathrm{ar}}^{\mathrm{n}âˆ’1}\\ â‡’{\text{T}}_{10}=500\left(\frac{11}{10}\right)Ã—{\left(\frac{11}{10}\right)}^{10âˆ’1}\\ \text{â€‰â€‰â€‰â€‰â€‰â€‰ â€‰}=500{\left(1.1\right)}^{10}\\ \mathrm{Thus},\text{}Ã¢â€šÂ¹\text{â€‰500 amounts to}Ã¢â€šÂ¹\text{â€‰}500{\left(1.1\right)}^{10}\text{in 10 years after its}\\ \text{deposit in a bank.}\end{array}$

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

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

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

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

$\frac{1}{1Ã—2}+\frac{1}{2Ã—3}+\frac{1}{3Ã—4}+\dots$

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

Ans

$\begin{array}{l}1.\text{}\mathrm{Let}{\text{â€‰ S}}_{\text{n}}=\text{1}Ã—\text{2}+\text{2}Ã—\text{3}+\text{3}Ã—\text{4}+\text{4}Ã—\text{5}+...+\mathrm{n}Ã—\left(\mathrm{n}+1\right)\\ \text{â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰}{\mathrm{a}}_{\mathrm{n}}=\mathrm{n}Ã—\left(\mathrm{n}+1\right)\\ \text{â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰}{\mathrm{S}}_{\mathrm{n}}=âˆ‘\mathrm{n}Ã—\left(\mathrm{n}+1\right)\\ \text{â€‰â€‰â€‰â€‰}=âˆ‘{\mathrm{n}}^{2}+âˆ‘\mathrm{n}\\ \text{â€‰â€‰â€‰â€‰}=\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{â€‰â€‰â€‰â€‰}=\frac{\mathrm{n}\left(\mathrm{n}+1\right)}{2}\left(\frac{2\mathrm{n}+1}{3}+1\right)\\ \text{â€‰â€‰â€‰â€‰}=\frac{\mathrm{n}\left(\mathrm{n}+1\right)}{2}\left(\frac{2\mathrm{n}+4}{3}\right)\\ \text{â€‰â€‰â€‰â€‰}=\frac{\mathrm{n}\left(\mathrm{n}+1\right)}{2}Ã—2\left(\frac{\mathrm{n}+2}{3}\right)\\ \text{â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰}{\mathrm{S}}_{\mathrm{n}}\text{â€‰}=\frac{1}{3}\mathrm{n}\left(\mathrm{n}+1\right)\left(\mathrm{n}+2\right)\\ \text{2.}\mathrm{Let}{\text{â€‰ S}}_{\text{n}}=\text{1}Ã—\text{2}Ã—\text{3}+\text{2}Ã—\text{3}Ã—\text{4}+\text{3}Ã—\text{4}Ã—\text{5}+...+\mathrm{n}Ã—\left(\mathrm{n}+1\right)Ã—\left(\mathrm{n}+2\right)\\ \text{â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰}{\mathrm{a}}_{\mathrm{n}}=\mathrm{n}Ã—\left(\mathrm{n}+1\right)Ã—\left(\mathrm{n}+2\right)\\ \text{â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰}{\mathrm{S}}_{\mathrm{n}}=âˆ‘\mathrm{n}Ã—\left(\mathrm{n}+1\right)Ã—\left(\mathrm{n}+2\right)\\ \text{â€‰â€‰â€‰â€‰}=âˆ‘\left({\mathrm{n}}^{3}+3{\mathrm{n}}^{2}+2\mathrm{n}\right)\\ \text{â€‰â€‰â€‰â€‰}=âˆ‘{\mathrm{n}}^{3}+3âˆ‘{\mathrm{n}}^{2}+2âˆ‘\mathrm{n}\\ \text{â€‰â€‰â€‰â€‰}={\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{â€‰â€‰â€‰â€‰}=\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{â€‰â€‰â€‰â€‰}=\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{â€‰â€‰â€‰â€‰}=\frac{\mathrm{n}\left(\mathrm{n}+1\right)}{2}\left\{\frac{{\mathrm{n}}^{2}+\mathrm{n}+4\mathrm{n}+2+4}{2}\right\}\\ \text{â€‰â€‰â€‰â€‰}=\frac{\mathrm{n}\left(\mathrm{n}+1\right)}{2}\left(\frac{{\mathrm{n}}^{2}+5\mathrm{n}+6}{2}\right)\\ \text{â€‰â€‰â€‰â€‰}=\frac{\mathrm{n}\left(\mathrm{n}+1\right)\left(\mathrm{n}+2\right)\left(\mathrm{n}+3\right)}{4}\\ \text{â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰}{\mathrm{S}}_{\mathrm{n}}\text{â€‰}=\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{â€‰}\mathrm{n}+1\right)Ã—{\mathrm{n}}^{2}\\ \text{â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰}{\mathrm{S}}_{\mathrm{n}}=âˆ‘\left(2{\mathrm{n}}^{3}+{\mathrm{n}}^{2}\right)\\ \text{â€‰â€‰â€‰â€‰}=2\text{â€‰}âˆ‘{\mathrm{n}}^{3}+âˆ‘{\mathrm{n}}^{2}\\ \text{â€‰â€‰â€‰â€‰}=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{â€‰â€‰â€‰â€‰}=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{â€‰â€‰â€‰â€‰}=\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{â€‰â€‰â€‰â€‰}=\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{â€‰â€‰â€‰â€‰}=\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{â€‰â€‰â€‰â€‰}=\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{â€‰â€‰â€‰â€‰}=\frac{\mathrm{n}\left(\mathrm{n}+1\right)}{2}\left\{\frac{3{\mathrm{n}}^{2}+5\mathrm{n}+1}{3}\right\}\\ \text{â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰}{\mathrm{S}}_{\mathrm{n}}\text{â€‰}=\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{â€‰â€‰}{\mathrm{S}}_{\mathrm{n}}=\frac{1}{1Ã—2}+\frac{1}{2Ã—3}+\frac{1}{3Ã—4}+...+\frac{1}{\mathrm{n}\left(\mathrm{n}+1\right)}\\ \text{â€‰â€‰â€‰â€‰}=\left(\frac{1}{1}âˆ’\frac{1}{2}\right)+\left(\frac{1}{2}âˆ’\frac{1}{3}\right)+\left(\frac{1}{3}âˆ’\frac{1}{4}\right)+...+\left(\frac{1}{\mathrm{n}}âˆ’\frac{1}{\mathrm{n}+1}\right)\\ \text{â€‰â€‰â€‰â€‰}=\text{â€‰}\frac{1}{1}âˆ’\frac{1}{\mathrm{n}+1}\\ \text{â€‰â€‰â€‰â€‰}=\frac{\mathrm{n}+1âˆ’1}{\mathrm{n}+1}\\ \text{â€‰â€‰â€‰â€‰}=\frac{\mathrm{n}}{\mathrm{n}+1}\\ \text{5.}{5}^{2}+{6}^{2}+{7}^{2}+...\text{}+\text{}{20}^{2}\\ \text{â€‰â€‰â€‰â€‰â€‰â€‰â€‰}={1}^{2}+{2}^{2}+{3}^{2}+{4}^{2}+{5}^{2}+{6}^{2}+{7}^{2}+...\text{}+\text{}{20}^{2}âˆ’\left({1}^{2}+{2}^{2}+{3}^{2}+{4}^{2}\right)\\ \text{â€‰â€‰â€‰â€‰â€‰â€‰â€‰}=âˆ‘{20}^{2}âˆ’âˆ‘{4}^{2}\\ \text{â€‰â€‰â€‰â€‰â€‰â€‰â€‰}=\frac{20\left(20+1\right)\left(2Ã—20+1\right)}{6}âˆ’\frac{4\left(4+1\right)\left(2Ã—4+1\right)}{6}\\ \text{â€‰â€‰â€‰â€‰â€‰â€‰â€‰}=\frac{20Ã—21Ã—41}{6}âˆ’\frac{4Ã—5Ã—9}{6}\\ \text{â€‰â€‰â€‰â€‰â€‰â€‰â€‰}=\frac{17220âˆ’180}{6}\\ \text{â€‰â€‰â€‰â€‰â€‰â€‰â€‰}=\frac{17040}{6}\\ \text{â€‰â€‰â€‰â€‰â€‰â€‰â€‰}=2840\\ 6.{\text{Letâ€‰S}}_{\text{n}}=\text{3}Ã—\text{8}+\text{6}Ã—\text{11}+\text{9}Ã—\text{14}+...\\ \text{â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰}{\mathrm{T}}_{\mathrm{n}}=\left(3,6,9,...{\mathrm{n}}^{\mathrm{th}}\text{â€‰}\mathrm{term}\right)\left(8,11,14,...{\text{n}}^{\text{th}}\text{term}\right)\\ \text{â€‰â€‰â€‰}=\left\{3+\left(\mathrm{n}âˆ’1\right)3\right\}\left\{8+\left(\mathrm{n}âˆ’1\right)3\right\}\\ \text{â€‰â€‰â€‰}=\left(3\text{â€‰}\mathrm{n}\right)\left(3\text{â€‰}\mathrm{n}+5\right)\\ \text{â€‰â€‰â€‰}=9\text{â€‰}{\mathrm{n}}^{2}+15\mathrm{n}\\ \text{â€‰â€‰â€‰â€‰â€‰â€‰â€‰}{\mathrm{S}}_{\mathrm{n}}=âˆ‘{\mathrm{T}}_{\mathrm{n}}\\ \text{â€‰â€‰â€‰â€‰}=âˆ‘\left(9{\mathrm{n}}^{2}+15\mathrm{n}\right)\\ \text{â€‰â€‰â€‰â€‰}=9âˆ‘{\mathrm{n}}^{2}+15âˆ‘\mathrm{n}\\ \text{â€‰â€‰â€‰â€‰}=9Ã—\frac{\mathrm{n}\left(\mathrm{n}+1\right)\left(2\mathrm{n}+1\right)}{6}+15Ã—\frac{\mathrm{n}\left(\mathrm{n}+1\right)}{2}\\ \text{â€‰â€‰â€‰â€‰}=3Ã—\frac{\mathrm{n}\left(\mathrm{n}+1\right)\left(2\mathrm{n}+1\right)}{2}+15Ã—\frac{\mathrm{n}\left(\mathrm{n}+1\right)}{2}\\ \text{â€‰â€‰â€‰â€‰}=\frac{3\mathrm{n}\left(\mathrm{n}+1\right)}{2}\left\{\left(2\mathrm{n}+1\right)+5\right\}\\ \text{â€‰â€‰â€‰â€‰}=\frac{3\mathrm{n}\left(\mathrm{n}+1\right)}{2}\left(2\mathrm{n}+6\right)\\ \text{â€‰â€‰â€‰â€‰}=3\mathrm{n}\left(\mathrm{n}+1\right)\left(\mathrm{n}+3\right)\\ 7.\text{}\mathrm{Let}\text{â€‰â€‰â€‰â€‰â€‰â€‰â€‰}{\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{â€‰â€‰}+\left({\text{1}}^{\text{2}}+{\text{2}}^{\text{2}}+{\text{3}}^{\text{2}}+...+{\mathrm{n}}^{2}\right)\\ \mathrm{Here},{\text{â€‰T}}_{\text{n}}={\text{1}}^{\text{2}}+{\text{2}}^{\text{2}}+{\text{3}}^{\text{2}}+...+{\mathrm{n}}^{2}\\ =âˆ‘{\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{â€‰â€‰â€‰â€‰}{\mathrm{S}}_{\mathrm{n}}=âˆ‘{\mathrm{T}}_{\mathrm{n}}\\ =\frac{1}{6}âˆ‘\left(2{\mathrm{n}}^{3}+3{\mathrm{n}}^{2}+\mathrm{n}\right)\\ =\frac{1}{6}\left(2âˆ‘{\mathrm{n}}^{3}+3âˆ‘{\mathrm{n}}^{2}+âˆ‘\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}Ã—\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}Ã—\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{â€‰â€‰â€‰â€‰}{\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 â€“ 1)2

Ans

1.

${\text{Letâ€‰â€‰T}}_{\text{n}}=\text{n}\left(\text{n}+\text{1}\right)\left(\text{n}+\text{4}\right)$ $\begin{array}{l}\text{â€‰â€‰â€‰â€‰}=\mathrm{n}\left({\mathrm{n}}^{2}+5\mathrm{n}+4\right)\\ \text{â€‰â€‰â€‰â€‰}={\mathrm{n}}^{3}+5{\mathrm{n}}^{2}+4\mathrm{n}\\ \text{â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰}{\mathrm{S}}_{\mathrm{n}}=âˆ‘{\mathrm{T}}_{\mathrm{n}}\\ \text{â€‰â€‰â€‰â€‰}=âˆ‘\left({\mathrm{n}}^{3}+5{\mathrm{n}}^{2}+4\mathrm{n}\right)\\ \text{â€‰â€‰â€‰â€‰}=âˆ‘{\mathrm{n}}^{3}+5âˆ‘{\mathrm{n}}^{2}+4âˆ‘\mathrm{n}\\ \text{â€‰â€‰â€‰â€‰}={\left\{\frac{\mathrm{n}\left(\mathrm{n}+1\right)}{2}\right\}}^{2}+\frac{5\text{â€‰}\mathrm{n}\left(\mathrm{n}+1\right)\left(2\mathrm{n}+1\right)}{6}+\frac{4\mathrm{n}\left(\mathrm{n}+1\right)}{2}\\ \text{â€‰â€‰â€‰â€‰}=\frac{\mathrm{n}\left(\mathrm{n}+1\right)}{2}\left\{\frac{\mathrm{n}\left(\mathrm{n}+1\right)}{2}+5.\frac{\left(2\mathrm{n}+1\right)}{3}+4\right\}\\ \text{â€‰â€‰â€‰â€‰}=\frac{\mathrm{n}\left(\mathrm{n}+1\right)}{2}\left\{\frac{3\mathrm{n}\left(\mathrm{n}+1\right)+10\left(2\mathrm{n}+1\right)+24}{6}\right\}\\ \text{â€‰â€‰â€‰â€‰}=\frac{\mathrm{n}\left(\mathrm{n}+1\right)}{2}\left(\frac{3{\mathrm{n}}^{2}+3\mathrm{n}+20\text{â€‰}\mathrm{n}+10+24}{6}\right)\\ {\mathrm{S}}_{\mathrm{n}}=\frac{\mathrm{n}\left(\mathrm{n}+1\right)}{12}\left(3{\mathrm{n}}^{2}+23\text{â€‰}\mathrm{n}+34\right)\end{array}$

2.

$\begin{array}{l}\mathrm{Let}{\text{â€‰â€‰T}}_{\text{n}}={\text{n}}^{\text{2}}+{\text{2}}^{\text{n}}\\ {\mathrm{S}}_{\mathrm{n}}=âˆ‘{\mathrm{T}}_{\mathrm{n}}\\ \text{â€‰â€‰â€‰â€‰â€‰â€‰â€‰}=âˆ‘\left({\text{n}}^{\text{2}}+{\text{2}}^{\text{n}}\right)\\ \text{â€‰â€‰â€‰â€‰â€‰â€‰â€‰}=âˆ‘{\text{n}}^{\text{2}}+âˆ‘{\text{2}}^{\text{n}}\\ \text{â€‰â€‰â€‰â€‰â€‰â€‰â€‰}=\frac{\mathrm{n}\left(\mathrm{n}+1\right)\left(2\mathrm{n}+1\right)}{6}+\left(2+{2}^{2}+{2}^{3}+{2}^{4}+...+{2}^{\mathrm{n}}\right)\end{array}$ $\begin{array}{l}\text{}\text{â€‰}\text{â€‰}\text{â€‰}\text{â€‰}\text{â€‰}\text{â€‰}\text{â€‰}\end{array}$ $\begin{array}{l}=\frac{\mathrm{n}\left(\mathrm{n}+1\right)\left(2\mathrm{n}+1\right)}{6}+\frac{2\left({2}^{\mathrm{n}}âˆ’1\right)}{2âˆ’1}\text{â€‰â€‰â€‰â€‰}\left[Ã¢Ë†Âµ{\mathrm{S}}_{\mathrm{n}}=\frac{\mathrm{a}\left({\mathrm{r}}^{\mathrm{n}}âˆ’1\right)}{\mathrm{r}âˆ’1}\right]\\ \text{â€‰â€‰â€‰â€‰â€‰â€‰â€‰}=\frac{\mathrm{n}\left(\mathrm{n}+1\right)\left(2\mathrm{n}+1\right)}{6}+2\left({2}^{\mathrm{n}}âˆ’1\right)\\ âˆ´{\mathrm{S}}_{\mathrm{n}}=\frac{1}{6}\mathrm{n}\left(\mathrm{n}+1\right)\left(2\mathrm{n}+1\right)+2\left({2}^{\mathrm{n}}âˆ’1\right)\end{array}$

3.

$\begin{array}{l}\mathrm{Let}\text{â€‰â€‰â€‰}{\mathrm{t}}_{\mathrm{n}}={\left(\text{2n}âˆ’\text{1}\right)}^{\text{2}}\\ \text{â€‰â€‰â€‰â€‰â€‰}=4{\mathrm{n}}^{2}âˆ’4\mathrm{n}+1\\ \text{â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰}{\mathrm{S}}_{\mathrm{n}}=âˆ‘\left(4{\mathrm{n}}^{2}âˆ’4\mathrm{n}+1\right)\\ \text{â€‰â€‰â€‰â€‰â€‰â€‰}=4âˆ‘{\mathrm{n}}^{2}âˆ’4âˆ‘\mathrm{n}+âˆ‘1\\ \text{â€‰â€‰â€‰â€‰â€‰â€‰}=4\frac{\mathrm{n}\left(\mathrm{n}+1\right)\left(2\mathrm{n}+1\right)}{6}âˆ’4\frac{\mathrm{n}\left(\mathrm{n}+1\right)}{2}+\mathrm{n}\\ \text{â€‰â€‰â€‰â€‰â€‰â€‰}=2\frac{\mathrm{n}\left(\mathrm{n}+1\right)\left(2\mathrm{n}+1\right)}{3}âˆ’2\mathrm{n}\left(\mathrm{n}+1\right)+\mathrm{n}\\ \text{â€‰â€‰â€‰â€‰â€‰â€‰}=\frac{2\mathrm{n}\left(2{\mathrm{n}}^{2}+3\text{â€‰}\mathrm{n}+1\right)}{3}âˆ’2{\mathrm{n}}^{2}âˆ’2\mathrm{n}+\mathrm{n}\\ \text{â€‰â€‰â€‰â€‰â€‰â€‰}=\frac{2\left(2{\mathrm{n}}^{3}+3\text{â€‰}{\mathrm{n}}^{2}+\mathrm{n}\right)}{3}âˆ’2{\mathrm{n}}^{2}âˆ’\mathrm{n}\\ \text{â€‰â€‰â€‰â€‰â€‰â€‰}=\frac{4{\mathrm{n}}^{3}+6\text{â€‰}{\mathrm{n}}^{2}+2\mathrm{n}âˆ’6{\mathrm{n}}^{2}âˆ’3\mathrm{n}}{3}\\ \text{â€‰â€‰â€‰â€‰â€‰â€‰}=\frac{4{\mathrm{n}}^{3}âˆ’\mathrm{n}}{3}\end{array}$ $\begin{array}{l}\text{â€‰â€‰â€‰â€‰â€‰â€‰}=\frac{\mathrm{n}}{3}\left(4{\mathrm{n}}^{2}âˆ’1\right)\\ {\mathrm{S}}_{\mathrm{n}}=\frac{\mathrm{n}}{3}\left(2\mathrm{n}âˆ’1\right)\left(2\mathrm{n}+1\right)\end{array}$

Q.58

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

Ans

$\begin{array}{l}1.\text{â€‰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}}_{âˆž}=\frac{\mathrm{a}}{1âˆ’\mathrm{r}}\\ =\frac{1}{1âˆ’\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},â€¦\\ \mathrm{a}=6,\text{}\\ \text{r}=\frac{1.2}{6}=0.2\\ {\mathrm{S}}_{âˆž}=\frac{\mathrm{a}}{1âˆ’\mathrm{r}}\\ =\frac{6}{1âˆ’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{â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰}\mathrm{a}=5,\text{}\\ \text{â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰r}=\frac{\left(\frac{20}{7}\right)}{5}=\frac{4}{7}\\ {\mathrm{S}}_{âˆž}=\frac{\mathrm{a}}{1âˆ’\mathrm{r}}\\ =\frac{5}{1âˆ’\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{âˆ’3}{4},\frac{3}{16},\frac{âˆ’3}{64},...\\ \mathrm{a}=\frac{âˆ’3}{4},\text{}\\ \text{â€‰r}=\frac{\left(\frac{3}{16}\right)}{\left(\frac{âˆ’3}{4}\right)}\\ =âˆ’\frac{1}{4}\\ {\mathrm{S}}_{âˆž}=\frac{\mathrm{a}}{1âˆ’\mathrm{r}}=\frac{\left(âˆ’\frac{3}{4}\right)}{1âˆ’\left(\frac{âˆ’1}{4}\right)}=âˆ’\frac{3}{4}Ã—\frac{4}{5}=âˆ’\frac{3}{5}\\ \text{5. L.H.S.=}{3}^{\frac{1}{2}}Ã—{3}^{\frac{1}{4}}Ã—{3}^{\frac{1}{8}}...\\ \text{â€‰}={3}^{\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+...}\\ \text{â€‰}={3}^{\left(\frac{\frac{1}{2}}{1âˆ’\frac{1}{2}}\right)}\left[Ã¢Ë†Âµ\mathrm{S}âˆž=\frac{\mathrm{a}}{1âˆ’\mathrm{r}}\right]\\ \text{â€‰}={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{â€‰â€‰ â€‰}=\frac{1}{1âˆ’\mathrm{a}}\left[{\mathrm{S}}_{âˆž}=\frac{\mathrm{a}}{1âˆ’\mathrm{r}}\right]\\ \text{and y}=1+\mathrm{b}+{\mathrm{b}}^{2}+...\\ \text{â€‰â€‰â€‰}=\frac{1}{1âˆ’\mathrm{b}}\\ \mathrm{L}.\mathrm{H}.\mathrm{S}.=1+\mathrm{ab}+{\mathrm{a}}^{2}{\mathrm{b}}^{2}+...\\ \text{â€‰â€‰â€‰ â€‰}=\frac{1}{1âˆ’\mathrm{ab}}\\ \mathrm{R}.\mathrm{H}.\mathrm{S}.=\frac{\mathrm{xy}}{\mathrm{x}+\mathrm{y}âˆ’1}\\ \text{â€‰â€‰â€‰ â€‰}=\frac{\left(\frac{1}{1âˆ’\mathrm{a}}\right)\left(\frac{1}{1âˆ’\mathrm{b}}\right)}{\left(\frac{1}{1âˆ’\mathrm{a}}\right)+\left(\frac{1}{1âˆ’\mathrm{b}}\right)âˆ’1}\\ \text{â€‰â€‰â€‰ â€‰}=\frac{1}{1âˆ’\mathrm{b}+1âˆ’\mathrm{a}âˆ’\left(1âˆ’\mathrm{a}\right)\left(1âˆ’\mathrm{b}\right)}\\ \text{â€‰â€‰â€‰â€‰}=\frac{1}{2âˆ’\mathrm{b}âˆ’\mathrm{a}âˆ’\left(1âˆ’\mathrm{b}âˆ’\mathrm{a}+\mathrm{ab}\right)}\\ \text{â€‰â€‰â€‰â€‰}=\frac{1}{2âˆ’\mathrm{b}âˆ’\mathrm{a}âˆ’1+\mathrm{b}+\mathrm{a}âˆ’\mathrm{ab}}\\ \text{â€‰â€‰â€‰â€‰}=\frac{1}{1âˆ’\mathrm{ab}}\\ \text{So,L.H.S.=R.H.S.}\end{array}$

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

Ans

$\begin{array}{l}\mathrm{Let}\text{first term of A.P. be a and common difference be d.}\\ \text{Then,}\\ {\text{T}}_{\text{m+n}}+{\text{T}}_{\text{m-n}}=\left\{\mathrm{a}+\left(\mathrm{m}+\mathrm{n}âˆ’1\right)\mathrm{d}\right\}+\left\{\mathrm{a}+\left(\mathrm{m}âˆ’\mathrm{n}âˆ’1\right)\mathrm{d}\right\}\\ \text{â€‰â€‰â€‰}=2\mathrm{a}+\left(\mathrm{m}+\mathrm{n}âˆ’1+\mathrm{m}âˆ’\mathrm{n}âˆ’1\right)\mathrm{d}\\ \text{â€‰â€‰â€‰}=2\mathrm{a}+2\left(\mathrm{m}âˆ’1\right)\mathrm{d}\\ \text{â€‰â€‰â€‰}=2\left\{\mathrm{a}+\left(\mathrm{m}âˆ’1\right)\mathrm{d}\right\}\\ \text{â€‰â€‰}=2{\mathrm{T}}_{\mathrm{m}}\\ \mathrm{Thus},\\ {\text{T}}_{\text{m+n}}+{\text{T}}_{\text{m-n}}=2{\mathrm{T}}_{\mathrm{m}}\\ \mathrm{Hence}\text{proved.}\end{array}$

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

Ans

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

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 â€“ S1).

Ans

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

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

Ans

$\begin{array}{l}\mathrm{The}\text{numbers between 200 to 400 divisible by 7 are:}\\ 203,\text{â€‰}210,\text{â€‰}217,224,\text{â€‰}231,...,\text{â€‰}399.\\ \mathrm{Here},\text{a}=203,\text{d}=210âˆ’203=7,\text{â€‰â€‰}\mathrm{l}=399\\ Ã¢Ë†Âµ\text{â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰}\mathrm{l}=\mathrm{a}+\left(\mathrm{n}âˆ’1\right)\mathrm{d}\\ \text{â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰}399=203+\left(\mathrm{n}âˆ’1\right)7\\ â‡’\text{â€‰}\frac{196}{7}=\mathrm{n}âˆ’1\end{array}$ $\begin{array}{l}â‡’28+1=\mathrm{n}\\ â‡’\text{â€‰â€‰â€‰â€‰â€‰}\mathrm{n}=29\\ \text{â€‰â€‰}{\mathrm{S}}_{\mathrm{n}}=\frac{\mathrm{n}}{2}\left(\mathrm{a}+\mathrm{l}\right)\\ â‡’\text{â€‰}{\mathrm{S}}_{29}=\frac{29}{2}\left(203+399\right)\\ \text{â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰}=\frac{29}{2}Ã—602\\ \text{â€‰â€‰â€‰â€‰â€‰â€‰â€‰ â€‰}=8729\\ \mathrm{Thus},\text{the sum of multiples of 7 between 200}\\ \text{and 400 is 8729.}\end{array}$

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

Ans

$\begin{array}{l}\mathrm{The}\text{numbers from 1 to 100 divisible by 2:}\\ \text{â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰}2,4,6,8,...,100\\ \mathrm{Here},\text{a}=2,\text{d}=4âˆ’2=2,\text{â€‰â€‰}\mathrm{l}=100\\ Ã¢Ë†Âµ\text{â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰}\mathrm{l}=\mathrm{a}+\left(\mathrm{n}âˆ’1\right)\mathrm{d}\\ \text{â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰}100=2+\left(\mathrm{n}âˆ’1\right)2\\ \text{â€‰â€‰â€‰}\frac{98}{2}=\mathrm{n}âˆ’1\\ â‡’\text{â€‰â€‰â€‰â€‰â€‰â€‰}\mathrm{n}=49+1\\ =50\\ Ã¢Ë†Âµ\text{â€‰â€‰}{\mathrm{S}}_{\mathrm{n}}=\frac{\mathrm{n}}{2}\left(\mathrm{a}+\mathrm{l}\right)\end{array}$ $\begin{array}{l}âˆ´{\mathrm{S}}_{50}=\frac{50}{2}\left(2+100\right)\\ =25Ã—102\\ =2550\\ \mathrm{The}\text{numbers from 1 to 100 divisible by 5:}\\ 5,10,15,20,...,100\\ \mathrm{Here},\text{a}=5,\text{d}=10âˆ’5=5,\text{â€‰â€‰}\mathrm{l}=100\\ Ã¢Ë†Âµ\text{â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰}\mathrm{l}=\mathrm{a}+\left(\mathrm{n}âˆ’1\right)\mathrm{d}\\ \text{â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰}100=5+\left(\mathrm{n}âˆ’1\right)5\\ \text{â€‰â€‰â€‰}\frac{95}{5}=\mathrm{n}âˆ’1\\ â‡’\text{â€‰â€‰â€‰â€‰â€‰â€‰}\mathrm{n}=19+1\\ =20\\ Ã¢Ë†Âµ\text{â€‰â€‰}{\mathrm{S}}_{\mathrm{n}}=\frac{\mathrm{n}}{2}\left(\mathrm{a}+\mathrm{l}\right)\\ âˆ´{\mathrm{S}}_{20}=\frac{20}{2}\left(5+100\right)\\ =10Ã—105\\ =1050\\ \mathrm{The}\text{numbers from 1 to 100 divisible by 2 and 5:}\\ \text{10,20,30,}...\text{,100}\\ \mathrm{Here},\text{a}=10,\text{d}=20âˆ’10=10,\text{â€‰â€‰}\mathrm{l}=100\\ Ã¢Ë†Âµ\text{â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰}\mathrm{l}=\mathrm{a}+\left(\mathrm{n}âˆ’1\right)\mathrm{d}\\ \text{â€‰â€‰â€‰â€‰â€‰â€‰â€‰}100=10+\left(\mathrm{n}âˆ’1\right)10\\ â‡’\text{â€‰â€‰â€‰}\frac{90}{10}=\mathrm{n}âˆ’1\end{array}$ $\begin{array}{l}\end{array}$ $\begin{array}{l}â‡’\text{â€‰â€‰â€‰â€‰â€‰}\mathrm{n}=10\\ Ã¢Ë†Âµ\text{â€‰â€‰}{\mathrm{S}}_{\mathrm{n}}=\frac{\mathrm{n}}{2}\left(\mathrm{a}+\mathrm{l}\right)\\ âˆ´{\mathrm{S}}_{10}=\frac{10}{2}\left(10+100\right)\\ \text{â€‰â€‰â€‰â€‰â€‰â€‰â€‰}=5Ã—110\\ \text{â€‰â€‰â€‰â€‰â€‰â€‰â€‰}=550\\ \mathrm{Sum}\text{of numbers divisible by 2 or 5}\\ =\mathrm{Sum}\text{of multiples of 2}+\mathrm{Sum}\text{of multiples of 5}\\ âˆ’\mathrm{Sum}\text{of multiples of 10}\\ =2550+1050âˆ’550\\ =3050\end{array}$

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

Ans

$\begin{array}{l}\mathrm{Two}\text{digit numbers which when divided by 4, yields}\\ \text{1 as remainder are:}\\ 13,\text{â€‰}17,\text{â€‰}21,...,\text{â€‰}97\\ \mathrm{a}=13,\text{â€‰}\mathrm{d}=17âˆ’13=4,\text{â€‰}\mathrm{l}=97\\ Ã¢Ë†Âµ\text{â€‰}\mathrm{l}=\mathrm{a}+\left(\mathrm{n}âˆ’1\right)\mathrm{d}\\ \text{â€‰â€‰â€‰â€‰}97=13+\left(\mathrm{n}âˆ’1\right)\mathrm{d}\\ 97âˆ’13=\left(\mathrm{n}âˆ’1\right)4\\ \text{â€‰â€‰â€‰â€‰â€‰â€‰}\frac{84}{4}=\mathrm{n}âˆ’1\\ \mathrm{n}=21+1\\ \text{â€‰â€‰â€‰ â€‰}=22\end{array}$ $\begin{array}{l}Ã¢Ë†Âµ\text{â€‰â€‰â€‰â€‰â€‰}{\mathrm{S}}_{\mathrm{n}}=\frac{\mathrm{n}}{2}\left(\mathrm{a}+\mathrm{l}\right)\\ âˆ´\text{â€‰â€‰â€‰}{\mathrm{S}}_{22}=\frac{22}{2}\left(13+97\right)\\ \text{â€‰â€‰â€‰â€‰}=\frac{22}{2}Ã—110\\ \text{â€‰â€‰â€‰â€‰}=1210\\ \mathrm{Thus},\text{the sum of 22 terms is 1210.}\end{array}$

Q.65

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

Ans

$\begin{array}{l}\mathrm{Since},\text{â€‰}\mathrm{f}\left(\mathrm{x}+\mathrm{y}\right)=\mathrm{f}\left(\mathrm{x}\right)\mathrm{f}\left(\mathrm{y}\right)\text{, f}\left(1\right)=3\text{â€‰and}\underset{\mathrm{x}=1}{\overset{\mathrm{n}}{âˆ‘}}\mathrm{f}\left(\mathrm{x}\right)=120.\\ \mathrm{So},\text{â€‰â€‰}\mathrm{f}\left(2\right)=\mathrm{f}\left(1+1\right)\\ =\mathrm{f}\left(1\right)\mathrm{f}\left(1\right)\\ =3Ã—3\\ ={3}^{2}\\ \text{â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰}\mathrm{f}\left(3\right)=\mathrm{f}\left(2+1\right)\\ =\mathrm{f}\left(2\right)\mathrm{f}\left(1\right)\\ ={3}^{2}Ã—3\\ ={3}^{3}\\ \text{â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰}\mathrm{f}\left(4\right)=\mathrm{f}\left(2+2\right)\\ =\mathrm{f}\left(2\right)\mathrm{f}\left(2\right)\end{array}$ $\begin{array}{l}={3}^{2}Ã—{3}^{2}\\ ={3}^{4}\\ Ã¢Ë†Âµ\underset{\mathrm{x}=1}{\overset{\mathrm{n}}{âˆ‘}}\mathrm{f}\left(\mathrm{x}\right)=120\\ â‡’\mathrm{f}\left(1\right)+\mathrm{f}\left(2\right)+\mathrm{f}\left(3\right)+\mathrm{f}\left(4\right)+...+\mathrm{f}\left(\mathrm{n}\right)=120\\ 3+{3}^{2}+{3}^{3}+{3}^{4}+...+{3}^{\mathrm{n}}=120\end{array}$ $\begin{array}{l}\end{array}$ $\begin{array}{l}â‡’\text{â€‰â€‰â€‰â€‰â€‰â€‰}\frac{3\left({3}^{\mathrm{n}}âˆ’1\right)}{3âˆ’1}=120\\ â‡’\text{â€‰â€‰â€‰â€‰â€‰â€‰â€‰ â€‰}{3}^{\mathrm{n}}âˆ’1=120Ã—\frac{2}{3}\\ â‡’\text{â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰}{3}^{\mathrm{n}}=80+1\\ â‡’\text{â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰}{3}^{\mathrm{n}}={3}^{4}\\ â‡’\text{â€‰}\mathrm{n}=4\\ \mathrm{Thus},\text{the value of n is 4.}\end{array}$

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

$\begin{array}{l}\mathrm{First}\text{term of G.P.}=\text{5}\\ \mathrm{Common}\text{difference of G.P.}=\text{2}\\ \text{Let number of terms in G.P.}=\text{n}\\ \text{â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰Sum of n terms in G.P.}=315\\ 5Ã—\frac{\left({2}^{\mathrm{n}}âˆ’1\right)}{2âˆ’1}=315\end{array}$ $\begin{array}{l}â‡’{2}^{\mathrm{n}}âˆ’1=\frac{315}{5}\\ {2}^{\mathrm{n}}=63+1\\ â‡’{2}^{\mathrm{n}}={2}^{6}\\ â‡’\mathrm{n}=6\\ Ã¢Ë†Âµ\mathrm{l}={\mathrm{ar}}^{\mathrm{n}âˆ’1}\\ \text{â€‰â€‰}=5Ã—{2}^{\left(6âˆ’1\right)}\\ \text{â€‰ â€‰}=5Ã—32\\ \text{â€‰â€‰}=160\\ \mathrm{Thus},\text{the last term of G.P. is 160 and number of}\\ \text{terms in G.P. is 6.}\end{array}$

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

$\begin{array}{l}\text{â€‰}\mathrm{First}\text{term of G.P.}=1\\ \mathrm{Let}\text{common ratio}=\mathrm{r}\\ \text{â€‰â€‰â€‰â€‰â€‰ â€‰}{\mathrm{T}}_{3}+{\mathrm{T}}_{5}=90\\ â‡’{\mathrm{ar}}^{2}+{\mathrm{ar}}^{4}=90\\ â‡’\text{â€‰â€‰}1.{\mathrm{r}}^{2}+1.{\mathrm{r}}^{4}=90\\ â‡’\text{â€‰â€‰â€‰â€‰â€‰â€‰â€‰}{\mathrm{r}}^{4}+{\mathrm{r}}^{2}=90\\ â‡’\text{â€‰}{\mathrm{r}}^{4}+{\mathrm{r}}^{2}âˆ’90=0\\ â‡’\text{â€‰â€‰}\left({\mathrm{r}}^{2}+10\right)\left({\mathrm{r}}^{2}âˆ’9\right)=0\\ â‡’\text{â€‰}{\mathrm{r}}^{2}=9,âˆ’10\end{array}$ $\begin{array}{l}â‡’\mathrm{r}=Â±3\left[\mathrm{Neglecting}{\text{r}}^{\text{2}}=âˆ’10\right]\\ \mathrm{Thus},\text{the common ratio of G.P. is}Â±3.\end{array}$

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

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

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

$\begin{array}{l}\mathrm{Let}\text{G.P. containing even number of terms is:}\\ {\text{a, ar, ar}}^{\text{2}}{\text{, ar}}^{\text{3}}\text{,}...{\text{, ar}}^{\text{n}},{\text{ar}}^{\text{n+1}},.\dots ,{\mathrm{ar}}^{2\mathrm{n}âˆ’1},{\mathrm{ar}}^{2\mathrm{n}}\\ \mathrm{According}\text{to given condition,}\\ {\mathrm{S}}_{2\mathrm{n}}=5Ã—\left(\mathrm{a}+{\mathrm{ar}}^{2}+{\mathrm{ar}}^{4}+...+{\mathrm{ar}}^{2\mathrm{n}âˆ’1}\right)\\ \left(\begin{array}{l}\text{a}+\text{ar}+{\text{ar}}^{\text{2}}+{\text{ar}}^{\text{3}}+...+{\text{ar}}^{\text{n}}\\ +{\text{ar}}^{\text{n+1}}+...+{\mathrm{ar}}^{2\mathrm{n}âˆ’1}+{\mathrm{ar}}^{2\mathrm{n}}\end{array}\right)=5Ã—\left(\mathrm{a}+{\mathrm{ar}}^{2}+{\mathrm{ar}}^{4}+...+{\mathrm{ar}}^{2\mathrm{n}âˆ’1}\right)\end{array}$ $\begin{array}{l}\left(\text{ar}+{\text{ar}}^{\text{3}}+...+{\text{ar}}^{\text{n}}+...+{\mathrm{ar}}^{2\mathrm{n}}\right)=4Ã—\left(\mathrm{a}+{\mathrm{ar}}^{2}+{\mathrm{ar}}^{4}+...+{\mathrm{ar}}^{2\mathrm{n}âˆ’1}\right)\\ \frac{\mathrm{ar}\left({\mathrm{r}}^{2\mathrm{n}}âˆ’1\right)}{\mathrm{r}âˆ’1}=4Ã—\frac{\mathrm{a}\left\{{\left({\mathrm{r}}^{2}\right)}^{\mathrm{n}}âˆ’1\right\}}{\mathrm{r}âˆ’1}\\ â‡’\mathrm{r}\left({\mathrm{r}}^{2\mathrm{n}}âˆ’1\right)=4\left({\mathrm{r}}^{2\mathrm{n}}âˆ’1\right)\\ â‡’\text{â€‰â€‰â€‰â€‰â€‰â€‰â€‰}\mathrm{r}=4\\ \mathrm{Thus},\text{the common ratio of G.P. is 4.}\end{array}$

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

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

Q.71

Ans

$\begin{array}{l}â‡’\frac{\mathrm{a}+\mathrm{bx}}{\mathrm{a}âˆ’\mathrm{bx}}=\frac{\mathrm{b}+\mathrm{cx}}{\mathrm{b}âˆ’\mathrm{cx}}\\ â‡’\left(\mathrm{a}+\mathrm{bx}\right)\left(\mathrm{b}âˆ’\mathrm{cx}\right)=\left(\mathrm{b}+\mathrm{cx}\right)\left(\mathrm{a}âˆ’\mathrm{bx}\right)\\ â‡’\mathrm{ab}âˆ’\mathrm{acx}+{\mathrm{b}}^{2}\mathrm{x}âˆ’{\mathrm{bcx}}^{2}=\mathrm{ba}âˆ’{\mathrm{b}}^{2}\mathrm{x}+\mathrm{cax}âˆ’{\mathrm{cbx}}^{2}\\ â‡’2{\mathrm{b}}^{2}\mathrm{x}=2\mathrm{acx}\\ â‡’\text{â€‰â€‰â€‰â€‰â€‰}{\mathrm{b}}^{2}=\mathrm{ac}\\ â‡’\text{â€‰â€‰â€‰â€‰ â€‰}\frac{\mathrm{b}}{\mathrm{a}}=\frac{\mathrm{c}}{\mathrm{b}}...\left(\mathrm{i}\right)\\ \mathrm{And}\text{â€‰}\frac{\mathrm{b}+\mathrm{cx}}{\mathrm{b}âˆ’\mathrm{cx}}=\frac{\mathrm{c}+\mathrm{dx}}{\mathrm{c}âˆ’\mathrm{dx}}\\ â‡’\left(\mathrm{b}+\mathrm{cx}\right)\left(\mathrm{c}âˆ’\mathrm{dx}\right)=\left(\mathrm{c}+\mathrm{dx}\right)\left(\mathrm{b}âˆ’\mathrm{cx}\right)\\ â‡’\mathrm{bc}âˆ’\mathrm{bdx}+{\mathrm{c}}^{2}\mathrm{x}âˆ’{\mathrm{cdx}}^{2}=\mathrm{cb}âˆ’{\mathrm{c}}^{2}\mathrm{x}+\mathrm{dbx}âˆ’{\mathrm{dcx}}^{2}\\ â‡’2{\mathrm{c}}^{2}\mathrm{x}=2\mathrm{bdx}\\ â‡’\text{â€‰â€‰â€‰â€‰â€‰}{\mathrm{c}}^{2}=\mathrm{bd}\\ â‡’\text{â€‰â€‰â€‰â€‰â€‰}\frac{\mathrm{c}}{\mathrm{b}}=\frac{\mathrm{d}}{\mathrm{c}}...\left(\mathrm{ii}\right)\\ \mathrm{From}\text{equation}\left(\mathrm{i}\right)\text{and equation}\left(\mathrm{ii}\right),\text{we get}\\ \frac{\mathrm{b}}{\mathrm{a}}=\frac{\mathrm{c}}{\mathrm{b}}=\frac{\mathrm{d}}{\mathrm{c}}\\ â‡’\mathrm{a},\mathrm{b},\mathrm{c},\mathrm{d}\text{are in G.P.}\end{array}$

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

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

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

Ans

$\begin{array}{l}\mathrm{Let}\text{first term of A.P. is A and common difference is D.}\\ \text{Then, according to given condition:}\\ \text{â€‰ â€‰â€‰â€‰â€‰â€‰}{\mathrm{T}}_{\mathrm{p}}=\mathrm{a}\\ â‡’\mathrm{A}+\left(\mathrm{p}âˆ’1\right)\mathrm{D}=\mathrm{a}...\left(\mathrm{i}\right)\\ \text{â€‰â€‰â€‰â€‰â€‰â€‰}{\mathrm{T}}_{\mathrm{q}}=\mathrm{b}\\ â‡’\mathrm{A}+\left(\mathrm{q}âˆ’1\right)\mathrm{D}=\mathrm{b}\text{â€‰}...\left(\mathrm{ii}\right)\\ \text{â€‰ â€‰â€‰â€‰â€‰â€‰}{\mathrm{T}}_{\mathrm{r}}=\mathrm{c}\\ â‡’\mathrm{A}+\left(\mathrm{r}âˆ’1\right)\mathrm{D}=\mathrm{c}\text{â€‰}...\left(\mathrm{iii}\right)\end{array}$ $\begin{array}{l}\mathrm{L}.\mathrm{H}.\mathrm{S}.=\left(\text{q}âˆ’\text{r}\right)\text{a}+\left(\text{r}âˆ’\text{p}\right)\text{b}+\left(\text{p}âˆ’\text{q}\right)\text{c}\\ =\left(\text{q}âˆ’\text{r}\right)\left\{\mathrm{A}+\left(\mathrm{p}âˆ’1\right)\mathrm{D}\right\}+\left(\text{r}âˆ’\text{p}\right)\left\{\mathrm{A}+\left(\mathrm{q}âˆ’1\right)\mathrm{D}\right\}\\ +\left(\text{p}âˆ’\text{q}\right)\left\{\mathrm{A}+\left(\mathrm{r}âˆ’1\right)\mathrm{D}\right\}\\ =\mathrm{A}\left(\text{q}âˆ’\text{r}+\text{r}âˆ’\text{p}+\text{p}âˆ’\text{q}\right)+\mathrm{D}\left\{\begin{array}{l}\left(\text{q}âˆ’\text{r}\right)\left(\mathrm{p}âˆ’1\right)+\left(\text{r}âˆ’\text{p}\right)\left(\mathrm{q}âˆ’1\right)\\ +\left(\text{p}âˆ’\text{q}\right)\left(\mathrm{r}âˆ’1\right)\end{array}\right\}\\ =\mathrm{A}\left(0\right)+\mathrm{D}\left(\mathrm{qp}âˆ’\mathrm{q}âˆ’\mathrm{rp}+\mathrm{r}+\mathrm{rq}âˆ’\mathrm{r}âˆ’\mathrm{pq}+\mathrm{p}+\mathrm{pr}âˆ’\mathrm{p}âˆ’\mathrm{qr}+\mathrm{q}\right)\\ =0+0\\ =0=\mathrm{R}.\mathrm{H}.\mathrm{S}.\\ \mathrm{This}\text{was to be proved.}\end{array}$

Q.74

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

Ans

$\begin{array}{l}\text{Since,a}\left(\frac{\text{1}}{\text{b}}\text{+}\frac{\text{1}}{\text{c}}\right)\text{,â€‰b}\left(\frac{\text{1}}{\text{c}}\text{+}\frac{\text{1}}{\text{a}}\right)\text{,c}\left(\frac{\text{1}}{\text{a}}\text{+}\frac{\text{1}}{\text{b}}\right)\text{areinA.P.}\\ â‡’\text{}\left(\frac{\text{a}}{\text{b}}\text{+}\frac{\text{a}}{\text{c}}\right)\text{,}\left(\frac{\text{b}}{\text{c}}\text{+}\frac{\text{b}}{\text{a}}\right)\text{,â€‰}\left(\frac{\text{c}}{\text{a}}\text{+}\frac{\text{c}}{\text{b}}\right)\text{areinA.P.}\\ â‡’\text{}\left(\frac{\text{a}}{\text{a}}\text{+}\frac{\text{a}}{\text{b}}\text{+}\frac{\text{a}}{\text{c}}\right)\text{,}\left(\frac{\text{b}}{\text{b}}\text{+}\frac{\text{b}}{\text{c}}\text{+}\frac{\text{b}}{\text{a}}\right)\text{,â€‰}\left(\frac{\text{c}}{\text{c}}\text{+}\frac{\text{c}}{\text{a}}\text{+}\frac{\text{c}}{\text{b}}\right)\text{areinA.P.}\\ â‡’\text{a}\left(\frac{\text{1}}{\text{a}}\text{+}\frac{\text{1}}{\text{b}}\text{+}\frac{\text{1}}{\text{c}}\right)\text{,b}\left(\frac{\text{1}}{\text{b}}\text{+}\frac{\text{1}}{\text{c}}\text{+}\frac{\text{1}}{\text{a}}\right)\text{,â€‰c}\left(\frac{\text{1}}{\text{c}}\text{+}\frac{\text{1}}{\text{a}}\text{+}\frac{\text{1}}{\text{b}}\right)\text{areinA.P.}\\ \mathrm{â‡’}\text{a,b,careinA.P.}\end{array}$

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{â€‰}\mathrm{a},\mathrm{b},\mathrm{c},\mathrm{d}\text{are in G}.\text{P}\\ â‡’\frac{\mathrm{b}}{\mathrm{a}}=\frac{\mathrm{c}}{\mathrm{b}}=\frac{\mathrm{d}}{\mathrm{c}}=\mathrm{r}\left(\mathrm{let}\right)\\ â‡’\mathrm{b}=\mathrm{ar},\mathrm{c}=\mathrm{br}\text{and d}=\text{cr}\\ â‡’\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{â€‰}{\mathrm{r}}^{2\mathrm{n}}\left(1+{\mathrm{r}}^{\mathrm{n}}\right)}{{\text{a}}^{\text{n}}\text{â€‰}{\mathrm{r}}^{\mathrm{n}}\left(\text{1}+{\mathrm{r}}^{\mathrm{n}}\right)}\\ ={\mathrm{r}}^{\mathrm{n}}\\ \mathrm{So},\text{â€‰â€‰â€‰}\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 â€“ 3x + p = 0 and c, d are roots of x2 â€“ 12x + q = 0, where a, b, c, d form a G.P. Prove that (q + p) : (q â€“ p) = 17:15.

Ans

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

Q.77

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

Ans

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

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

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

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

Ans

$\begin{array}{l}\left(\mathrm{i}\right)\mathrm{Let}\text{â€‰}{\mathrm{S}}_{\mathrm{n}}=\text{5}+\text{55}+\text{555}+â€¦\mathrm{}\\ \text{â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰ â€‰}=5\left(1+11+111+1111+...\right)\\ \text{â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰ â€‰â€‰â€‰â€‰â€‰â€‰â€‰}=\frac{5}{9}\left(9+99+999+9999+...\right)\\ \text{â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰}=\frac{5}{9}\left\{\left(10âˆ’1\right)+\left({10}^{2}âˆ’1\right)+\left({10}^{3}âˆ’1\right)+\left({10}^{4}âˆ’1\right)+...\right\}\\ \text{â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰ â€‰}=\frac{5}{9}\left\{10+{10}^{2}+{10}^{3}+{10}^{4}+...âˆ’1âˆ’1âˆ’1âˆ’1âˆ’...\text{â€‰}\mathrm{n}\text{times}\right\}\\ \text{â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰ â€‰}=\frac{5}{9}\left\{\frac{10\left({10}^{\mathrm{n}}âˆ’1\right)}{10âˆ’1}âˆ’\text{â€‰}\mathrm{n}\right\}\left[Ã¢Ë†Âµ{\mathrm{S}}_{\mathrm{n}}=\frac{\mathrm{a}\left({\mathrm{r}}^{\mathrm{n}}âˆ’1\right)}{\mathrm{r}âˆ’1}\right]\\ \text{â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰}=\frac{5}{9}\left\{\frac{10\left({10}^{\mathrm{n}}âˆ’1\right)}{9}âˆ’\text{â€‰}\mathrm{n}\right\}\\ \text{â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰ â€‰}=\frac{50}{81}\left({10}^{\mathrm{n}}âˆ’1\right)âˆ’\frac{5\mathrm{n}}{9}\\ \left(\mathrm{ii}\right)\mathrm{Let}\text{â€‰}{\mathrm{S}}_{\mathrm{n}}=0.\text{6}+0.\text{66}+0.\text{666}+â€¦\\ \text{â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰ â€‰}=\text{6}\left(0.1+0.11+111+1111+...\right)\\ \text{â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰}=\frac{\text{6}}{9}\left(0.9+0.99+0.999+0.9999+...\right)\\ \text{â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰}=\frac{\text{2}}{3}\left\{\left(1âˆ’0.1\right)+\left(1âˆ’0.01\right)+\left(1âˆ’0.001\right)+\left(1âˆ’0.0001\right)+...\right\}\\ \text{â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰ â€‰}=\frac{\text{2}}{3}\left\{1+1+1+1+...\text{â€‰}\mathrm{n}\text{times}âˆ’\frac{1}{10}âˆ’\frac{1}{{10}^{2}}âˆ’\frac{1}{{10}^{3}}âˆ’\frac{1}{{10}^{4}}âˆ’...\right\}\end{array}$ $\begin{array}{l}\text{â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰ â€‰}=\frac{\text{2}}{3}\left\{\mathrm{n}âˆ’\frac{\frac{1}{10}\left(1âˆ’\frac{1}{{10}^{\mathrm{n}}}\right)}{1âˆ’\frac{1}{10}}\right\}\left[Ã¢Ë†Âµ{\mathrm{S}}_{\mathrm{n}}=\frac{\mathrm{a}\left({\mathrm{r}}^{\mathrm{n}}âˆ’1\right)}{\mathrm{r}âˆ’1}\right]\\ \text{â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰ â€‰}=\frac{\text{2}}{3}\mathrm{n}âˆ’\frac{\text{2}}{3}Ã—\frac{\left(1âˆ’\frac{1}{{10}^{\mathrm{n}}}\right)}{9}\\ \text{â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰ â€‰}=\frac{\text{2}}{3}\mathrm{n}âˆ’\frac{\text{2}}{27}Ã—\left(1âˆ’\frac{1}{{10}^{\mathrm{n}}}\right)\end{array}$

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

Ans

$\begin{array}{l}\text{2}Ã—\text{4}+\text{4}Ã—\text{6}+\text{6}Ã—\text{8}+\text{}...\text{}+\text{nâ€‰terms}\\ {\mathrm{T}}_{\mathrm{n}}=\left\{2+\left(\mathrm{n}âˆ’1\right)2\right\}\left\{4+\left(\mathrm{n}âˆ’1\right)2\right\}\\ =\left(2\mathrm{n}\right)\left(2\mathrm{n}+2\right)\\ =4\mathrm{n}\left(\mathrm{n}+1\right)\\ \mathrm{Putting}\text{n}=20,\text{we get}\\ {\mathrm{T}}_{20}=4\left(20\right)\left(20+1\right)\\ =80Ã—21\\ =1680\\ \mathrm{Thus},{\text{the 20}}^{\text{th}}\text{term of series is 1680.}\end{array}$

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

Ans

$\begin{array}{l}\mathrm{Let}{\text{S}}_{\text{n}}=3+7+13+21+31+...+{\mathrm{a}}_{\mathrm{n}âˆ’1}+{\mathrm{a}}_{\mathrm{n}}\\ \mathrm{and}âˆ’\underset{Â¯}{{\text{S}}_{\text{n}}=\text{â€‰â€‰â€‰â€‰}âˆ’\text{â€‰}3âˆ’7âˆ’13âˆ’21âˆ’31âˆ’...âˆ’{\mathrm{a}}_{\mathrm{n}âˆ’1}âˆ’{\mathrm{a}}_{\mathrm{n}}}\\ \text{â€‰â€‰}0=3+\left\{4+6+8+10+..\text{â€‰}.\text{â€‰}\left(\mathrm{n}âˆ’1\right)\mathrm{terms}\right\}âˆ’\text{â€‰}{\mathrm{a}}_{\mathrm{n}}\\ {\mathrm{a}}_{\mathrm{n}}=3+\frac{\mathrm{n}âˆ’1}{2}\left\{8+\left(\mathrm{n}âˆ’2\right)2\right\}\\ \text{â€‰â€‰â€‰â€‰â€‰â€‰}=3+\left(\mathrm{n}âˆ’1\right)\left(4+\mathrm{n}âˆ’2\right)\\ \text{â€‰â€‰â€‰â€‰â€‰â€‰}=3+\left(\mathrm{n}âˆ’1\right)\left(\mathrm{n}+2\right)\\ {\mathrm{a}}_{\mathrm{n}}=3+{\mathrm{n}}^{2}+\mathrm{n}âˆ’2\\ \text{â€‰â€‰â€‰â€‰â€‰â€‰}={\mathrm{n}}^{2}+\mathrm{n}+1\\ \text{â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰â€‰}{\mathrm{S}}_{\mathrm{n}}=âˆ‘{\mathrm{n}}^{2}+âˆ‘\mathrm{n}+âˆ‘1\\ \text{â€‰â€‰â€‰â€‰â€‰}=\frac{\mathrm{n}\left(\mathrm{n}+1\right)\left(2\mathrm{n}+1\right)}{6}+\frac{\mathrm{n}\left(\mathrm{n}+1\right)}{2}+\mathrm{n}\\ \text{â€‰â€‰â€‰â€‰â€‰}=\frac{\mathrm{n}}{6}\left\{\left(\mathrm{n}+1\right)\left(2\mathrm{n}+1\right)+3\left(\mathrm{n}+1\right)+6\right\}\\ \text{â€‰â€‰â€‰â€‰â€‰}=\frac{\mathrm{n}}{6}\left(2{\mathrm{n}}^{2}+3\mathrm{n}+1+3\mathrm{n}+3+6\right)\\ \text{â€‰â€‰â€‰â€‰â€‰}=\frac{\mathrm{n}}{6}\left(2{\mathrm{n}}^{2}+6\mathrm{n}+10\right)\\ {\mathrm{S}}_{\mathrm{n}}=\frac{\mathrm{n}}{3}\left({\mathrm{n}}^{2}+3\mathrm{n}+5\right)\end{array}$

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

$\begin{array}{l}\mathrm{We}\text{are given that}\\ {\text{S}}_{\text{1}}=âˆ‘\mathrm{n}\\ \text{â€‰â€‰â€‰}=\frac{\mathrm{n}\left(\mathrm{n}+1\right)}{2}\\ {\text{S}}_{\text{2}}=âˆ‘{\mathrm{n}}^{2}\\ \text{â€‰â€‰ â€‰}=\frac{\mathrm{n}\left(\mathrm{n}+1\right)\left(2\mathrm{n}+1\right)}{6}\\ {\text{S}}_{\text{3}}=âˆ‘{\mathrm{n}}^{3}\\ \text{â€‰â€‰ â€‰}={\left\{\frac{\mathrm{n}\left(\mathrm{n}+2\right)}{2}\right\}}^{2}\\ \mathrm{R}.\mathrm{H}.\mathrm{S}.={\mathrm{S}}_{3}\left(1+8{\mathrm{S}}_{1}\right)\\ \text{â€‰â€‰â€‰}={\left\{\frac{\mathrm{n}\left(\mathrm{n}+2\right)}{2}\right\}}^{2}\left\{1+8Ã—\frac{\mathrm{n}\left(\mathrm{n}+1\right)}{2}\right\}\\ \text{â€‰â€‰ â€‰}=\frac{{\mathrm{n}}^{2}{\left(\mathrm{n}+2\right)}^{2}}{4}\left\{1+4\mathrm{n}\left(\mathrm{n}+1\right)\right\}\\ \text{â€‰â€‰ â€‰}=\frac{{\mathrm{n}}^{2}{\left(\mathrm{n}+2\right)}^{2}}{4}\left(1+4{\mathrm{n}}^{2}+4\mathrm{n}\right)\\ \text{â€‰â€‰ â€‰}=\frac{{\mathrm{n}}^{2}{\left(\mathrm{n}+2\right)}^{2}}{4}\left(4{\mathrm{n}}^{2}+4\mathrm{n}+1\right)\\ \text{â€‰â€‰â€‰}=\frac{{\mathrm{n}}^{2}{\left(\mathrm{n}+2\right)}^{2}{\left(2\mathrm{n}+1\right)}^{2}}{4}\\ \text{â€‰ â€‰â€‰}={\left\{3Ã—\frac{\mathrm{n}\left(\mathrm{n}+2\right)\left(2\mathrm{n}+1\right)}{6}\right\}}^{2}\end{array}$ $\begin{array}{l}\text{â€‰ â€‰â€‰}=9{\left\{\frac{\mathrm{n}\left(\mathrm{n}+2\right)\left(2\mathrm{n}+1\right)}{6}\right\}}^{2}\\ =9\text{â€‰}{\mathrm{S}}_{2}^{2}=\mathrm{L}.\mathrm{H}.\mathrm{S}.\\ \mathrm{This}\text{was to be proved.}\end{array}$

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

$\begin{array}{l}\frac{{1}^{3}}{1}+\frac{{1}^{3}+{2}^{3}}{1+3}+\frac{{1}^{3}+{2}^{3}+{3}^{3}}{1+3+5}+...\\ {\mathrm{T}}_{\mathrm{n}}=\frac{{1}^{3}+{2}^{3}+{3}^{3}+...+{\mathrm{n}}^{3}}{1+3+5+...}\\ =\frac{âˆ‘{\mathrm{n}}^{3}}{\frac{\mathrm{n}}{2}\left\{2Ã—1+\left(\mathrm{n}âˆ’1\right)2\right\}}\\ =\frac{{\left\{\frac{\mathrm{n}\left(\mathrm{n}+1\right)}{2}\right\}}^{2}}{\frac{\mathrm{n}}{2}\left(2+2\mathrm{n}âˆ’2\right)}\\ =\frac{{\mathrm{n}}^{2}{\left(\mathrm{n}+1\right)}^{2}}{4{\mathrm{n}}^{2}}\\ =\frac{1}{4}\left({\mathrm{n}}^{2}+2\mathrm{n}+1\right)\end{array}$ $\begin{array}{l}{\mathrm{S}}_{\mathrm{n}}=\frac{1}{4}\left(âˆ‘{\mathrm{n}}^{2}+2âˆ‘\mathrm{n}+âˆ‘1\right)\\ =\frac{1}{4}\left[\frac{\mathrm{n}\left(\mathrm{n}+1\right)\left(2\mathrm{n}+1\right)}{6}+\frac{2\mathrm{n}\left(\mathrm{n}+1\right)}{2}+\mathrm{n}\right]\\ =\frac{1}{4}\left\{\frac{\mathrm{n}\left(\mathrm{n}+1\right)\left(2\mathrm{n}+1\right)}{6}+\mathrm{n}+1+1\right\}\\ =\frac{\mathrm{n}}{4}\left(\frac{2{\mathrm{n}}^{2}+3\mathrm{n}+1+6\mathrm{n}+12}{6}\right)\\ =\frac{\mathrm{n}}{4}\left(\frac{2{\mathrm{n}}^{2}+9\mathrm{n}+13}{6}\right)\\ =\frac{\mathrm{n}}{24}\left(2{\mathrm{n}}^{2}+9\mathrm{n}+13\right)\end{array}$

Q.84

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

Ans

$\begin{array}{l}\mathrm{L}.\mathrm{H}.\mathrm{S}.=\frac{1Ã—{2}^{2}+2Ã—{3}^{2}+...+\mathrm{n}Ã—{\left(\mathrm{n}+1\right)}^{2}}{{1}^{2}Ã—2+{2}^{2}Ã—3+...+{\mathrm{n}}^{2}Ã—\left(\mathrm{n}+1\right)}\\ \text{â€‰â€‰}=\frac{âˆ‘\left\{\mathrm{n}Ã—{\left(\mathrm{n}+1\right)}^{2}\right\}}{âˆ‘\left\{{\mathrm{n}}^{2}Ã—\left(\mathrm{n}+1\right)\right\}}\\ \text{â€‰â€‰}=\frac{âˆ‘\left({\mathrm{n}}^{3}+2{\mathrm{n}}^{2}+\mathrm{n}\right)}{âˆ‘\left({\mathrm{n}}^{3}+{\mathrm{n}}^{2}\right)}\end{array}$ $\begin{array}{l}\text{â€‰â€‰}=\frac{âˆ‘{\mathrm{n}}^{3}+2âˆ‘{\mathrm{n}}^{2}+âˆ‘\mathrm{n}}{âˆ‘{\mathrm{n}}^{3}+âˆ‘{\mathrm{n}}^{2}}\\ \text{â€‰â€‰}=\frac{{\left\{\frac{\mathrm{n}\left(\mathrm{n}+1\right)}{2}\right\}}^{2}+2\left\{\frac{\mathrm{n}\left(\mathrm{n}+1\right)\left(2\mathrm{n}+1\right)}{6}\right\}+\frac{\mathrm{n}\left(\mathrm{n}+1\right)}{2}}{{\left\{\frac{\mathrm{n}\left(\mathrm{n}+1\right)}{2}\right\}}^{2}+\frac{\mathrm{n}\left(\mathrm{n}+1\right)\left(2\mathrm{n}+1\right)}{6}}\\ \text{â€‰â€‰}=\frac{\frac{\mathrm{n}\left(\mathrm{n}+1\right)}{2}\left\{\frac{\mathrm{n}\left(\mathrm{n}+1\right)}{2}+2Ã—\frac{2\mathrm{n}+1}{3}+1\right\}}{\frac{\mathrm{n}\left(\mathrm{n}+1\right)}{2}\left\{\frac{\mathrm{n}\left(\mathrm{n}+1\right)}{2}+\frac{2\mathrm{n}+1}{3}\right\}}\\ \text{â€‰â€‰}=\frac{\left(\frac{3{\mathrm{n}}^{2}+3\mathrm{n}+8\mathrm{n}+4+6}{6}\right)}{\left(\frac{3{\mathrm{n}}^{2}+3\mathrm{n}+4\mathrm{n}+2}{6}\right)}\\ =\frac{3{\mathrm{n}}^{2}+11\mathrm{n}+10}{3{\mathrm{n}}^{2}+7\mathrm{n}+2}\\ =\frac{\left(\mathrm{n}+2\right)\left(3\mathrm{n}+5\right)}{\left(\mathrm{n}+2\right)\left(3\mathrm{n}+1\right)}\end{array}$ $\begin{array}{l}\text{}\end{array}$ $\begin{array}{l}=\frac{\left(3\mathrm{n}+5\right)}{\left(3\mathrm{n}+1\right)}=\mathrm{R}.\mathrm{H}.\mathrm{S}.\\ \mathrm{This}\text{was to be proved.}\end{array}$

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

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