NCERT Solutions Class 11 Maths Chapter 9

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

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

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

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

Key Topics Covered In Chapter 9 Class 11 Mathematics

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

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

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

Introduction

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

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

Sequences

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

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

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

There are two types of sequences:

Infinite sequence

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

Finite sequence

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

Fibonacci sequence :

Example-

A1 = A2 = 1,

A3 = A1 + A2,

An = An-2 + An-1,

n > 2

Series

In this section, we will learn about the series.

There is an example given below that shows the series:

a1 + a2 + a3 + a4 + …… an + …..

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

The important points you should note about the series are:

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

Arithmetic progression (A.P)

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

a1 + a2 + a3 + a4 + …… an + …..

an+1 = an + d

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

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

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

The series given below follows arithmetic progression:

a, a + d, a + 2d, a + 3d…a + (n + 1)d

Formulas,

The last number of A.P,

l = a + (n – 1)d

The sum of n numbers in A.P,

Sn = n/2 [2a + (n – 1)d]

Sn = n/2 [a + l]

Arithmetic mean

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

Geometric progressions (G.P)

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

There is a series given below which follows arithmetic progression:

a1 + a2 + a3 + a4 + …… an + …..

ak+1 / ak = r (constant), for k ≥ 1

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

The series given below follows a Geometric progression:

a, ar, ar2, ar3,…..

Formulas,

General ‘n’ term of a G.P,

an = arn-1

The sum of n numbers in G.P,

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

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

Geometric mean

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

Relationship between A.M and G.M

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

A.M = (a + b) /2

G.M = √a.b

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

This is the relationship between A.M ≥ G.M

Sum to ‘n’ terms of special series.

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

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

Sn = n(n + 1) / 2

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

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

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

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

NCERT Solutions for Class 11 Mathematics Chapter 9 Exercise & Solutions

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

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

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

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

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NCERT Exemplar Class 11 Mathematics

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

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

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

Key Features of NCERT Solutions for Class 11 Mathematics Chapter 9

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

The key features are as follows:

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

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

Ans

$\begin{array}{l} We have \\ a_{n} = n (n + 2) \\ Substituting n = 1, 2, 3, 4, 5 respectively, we get \\ a_{1} = 1 (1 + 2) \\ = 3 \\ a_{2} = 2 (2 + 2) \\ = 8 \\ a_{3} = 3 (3 + 2) \\ = 15 \\ a_{4} = 4 (4 + 2) \\ = 24 \\ a_{5} = 5 (5 + 2) \\ = 35 \\ Therefore, the first five terms are: 3, 8, 15, 24, 35. \end{array}$

Q.2

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

Ans

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

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

Ans

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

Q.4

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

Ans

$\begin{array}{l} We have \\ a_{n} = \frac{2 n - 3}{6} \\ Substituting n = 1, 2, 3, 4, 5 respectively, we get \\ a_{1} = \frac{2 (1) - 3}{6} \\ = - \frac{1}{6} \\ a_{2} = \frac{2 (2) - 3}{6} \\ = \frac{1}{6} \end{array}$ $\begin{array}{l} a_{3} = \frac{2 (3) - 3}{6} \\ = \frac{3}{6} \\ = \frac{1}{2} \\ a_{4} = \frac{2 (4) - 3}{6} \\ = \frac{5}{6} \\ a_{5} = \frac{2 (5) - 3}{6} \\ = \frac{7}{6} \\ Therefore, the first five terms are: - \frac{1}{6}, \frac{1}{6}, \frac{1}{2}, \frac{5}{6}, \frac{7}{6} . \end{array}$

Q.5 Write the first five terms of the sequence whose nth terms is a_n = (– 1)^n–15ⁿ⁺¹.

Ans

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

Q.6

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

Ans

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

Q.7

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

Ans

$\begin{array}{l} Given: \\ a_{n} = 4n - 3 \\ Substituting n=17 and 24 respectively, we get \\ a_{17} = 4 (17) - 3 \\ = 68 - 3 \\ = 65 \\ a_{24} = 4 (24) - 3 \\ = 96 - 3 \\ = 93 \end{array}$

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

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

Q.8

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

Ans

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

$\begin{array}{l} Given : a_{1} = - 1, a_{n} = \frac{a_{n - 1}}{n}, n \geq 2 \\ Substituting n = 2, we get \\ a_{2} = \frac{a_{2 - 1}}{2} \\ = \frac{a_{1}}{2} \\ = \frac{- 1}{2} [Putting a_{1} = - 1] \\ Substituting n = 3, we get \\ a_{3} = \frac{a_{3 - 1}}{3} \\ = \frac{a_{2}}{3} \\ = \frac{(\frac{- 1}{2})}{3} [Putting a_{2} = \frac{- 1}{2}] \end{array}$ $\begin{array}{l} \end{array}$ $\begin{array}{l} = \frac{- 1}{6} \\ Substituting n = 4, we get \\ a_{4} = \frac{a_{4 - 1}}{4} \\ = \frac{a_{3}}{4} \end{array}$ $\begin{array}{l} \end{array}$ $\begin{array}{l} = \frac{(\frac{- 1}{6})}{4} [Putting a_{3} = \frac{- 1}{6}] \\ = \frac{- 1}{24} \\ Substituting n = 5, we get \\ a_{5} = \frac{a_{5 - 1}}{5} \\ = \frac{a_{4}}{5} \\ = \frac{(\frac{- 1}{24})}{5} [Putting a_{4} = \frac{- 1}{24}] \\ = \frac{- 1}{120} \\ Thus, the first five terms are - 1, \frac{- 1}{2}, \frac{- 1}{6}, \frac{- 1}{24} and \frac{- 1}{120} . \\ The corresponding series is: \\ (- 1) + (\frac{- 1}{2}) + (\frac{- 1}{6}) + (\frac{- 1}{24}) + (\frac{- 1}{120}) + . .. \end{array}$

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

Q.9

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

Ans

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

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

Ans

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

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

Ans

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

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 20^th term is –112.

Ans

$\begin{array}{l} First term of A.P. = 2 \\ Let common difference = d \\ According to given condition: \\ Sum of first 5 terms = \frac{1}{4} (Sum of next 5 terms) \\ S_{5} = \frac{1}{4} (S_{10} - S_{5}) \\ \Rightarrow 4 S_{5} = S_{10} - S_{5} \\ \Rightarrow 5 S_{5} = S_{10} \end{array}$ $\begin{array}{l} \Rightarrow 5 [\frac{5}{2} {2 \times 2 + (5 - 1) d}] = [\frac{10}{2} {2 \times 2 + (10 - 1) d}] \\ [âˆµ S_{n} = \frac{n}{2} {2 a + (n - 1) d}] \\ \Rightarrow \frac{25}{2} (4 + 4 d) = 5 (4 + 9 d) \\ \Rightarrow 25 (2 + 2 d) = 5 (4 + 9 d) \\ \Rightarrow 50 + 50 d = 20 + 45 d \\ \Rightarrow 5 d = 20 - 50 \\ \Rightarrow d = - \frac{30}{5} \\ = - 6 \\ Now, sum of 20 terms. \\ T_{20} = 2 + (20 - 1) \times - 6 [âˆµ a_{n} = a + (n - 1) d] \\ = 2 - 114 \\ = - 112 \\ Thus, {the 20}^{th} term is - 112. \end{array}$

Q.13

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

Ans

$\begin{array}{l} The given A.P. is - 6, - \frac{11}{2}, - 5, . .. \\ Let the sum of n terms of A.P. is - 25.Then, \end{array}$ $\begin{array}{l} S_{n} = \frac{n}{2} {2 a + (n - 1) d} \\ a = - 6 \\ d = - \frac{11}{2} - (- 6) \\ = - \frac{11}{2} + 6 = \frac{1}{2} \\ - 25 = \frac{n}{2} {2 \times - 6 + (n - 1) \times \frac{1}{2}} [âˆµ S_{n} = \frac{n}{2} {2 a + (n - 1) d}] \\ - 25 = \frac{n}{2} {- 12 + (n - 1) \times \frac{1}{2}} \\ - 100 = n (- 24 + n - 1) \\ - 100 = - 25 n + n^{2} \\ n^{2} - 25 n + 100 = 0 \\ (n - 20) (n - 5) = 0 \\ \Rightarrow n = 20, 5 \\ Thus, the number of terms of A.P. is 5 or 20. \end{array}$

Q.14

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

Ans

$\begin{array}{l} Let first term of A.P. = a \\ Let common difference of A.P. = d \\ Since, T_{p} = \frac{1}{q} \end{array}$ $\begin{array}{l} \Rightarrow a + (p - 1) d = \frac{1}{q} . .. (i) \\ And, T_{q} = \frac{1}{p} \\ \Rightarrow a + (q - 1) d = \frac{1}{p} . .. (ii) \\ Subtracting equation (ii) from equation (i), we get \\ (p - 1 - q + 1) d = \frac{1}{q} - \frac{1}{p} \\ \Rightarrow (p - q) d = \frac{p - q}{pq} \\ \Rightarrow d = \frac{1}{pq} \\ Substituting value of d in equation (i), we get \\ a + (p - 1) \frac{1}{pq} = \frac{1}{q} \\ \Rightarrow a + \frac{p}{pq} - \frac{1}{pq} = \frac{1}{q} \\ \Rightarrow a = \frac{1}{pq} \\ Now, S_{pq} = \frac{pq}{2} {2 \times \frac{1}{pq} + (pq - 1) \frac{1}{pq}} \\ [âˆµ S_{n} = \frac{n}{2} {2 a + (n - 1) d}] \end{array}$ $\begin{array}{l} \end{array}$ $\begin{array}{l} S_{pq} = \frac{pq}{2} {2 \times \frac{1}{pq} + (pq - 1) \frac{1}{pq}} \\ = \frac{pq}{2} (\frac{2}{pq} + 1 - \frac{1}{pq}) \\ = \frac{pq}{2} (\frac{1}{pq} + 1) \\ = \frac{1}{2} (1 + pq) \\ Thus, the sum of pq terms is \frac{1}{2} (pq + 1) . \end{array}$

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

Ans

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

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

Ans

$\begin{array}{l} k^{th} term of A.P. = 5k+1 \\ \Rightarrow T_{k} = 5k+1 \\ Substituting k = 1, 2, 3, . .. respectively. \\ T_{1} = 5 (1) + 1 \\ = 6 \\ T_{2} = 5 (2) + 1 \\ = 11 \\ T_{3} = 5 (3) + 1 \\ = 16 \\ Common difference = T_{2} - T_{1} \\ = 11 - 6 \\ = 5 \end{array}$ $\begin{array}{l} Sum {of n terms,S}_{n} = \frac{n}{2} {2 a + (n - 1) d} \\ = \frac{n}{2} {2 \times 6 + (n - 1) 5} \\ = \frac{n}{2} (12 + 5 n - 5) \\ = \frac{n}{2} (7 + 5 n) \\ = \frac{n}{2} (5 n + 7) \\ Thus, the sum of n terms is \frac{n}{2} (5 n + 7) . \end{array}$

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

$\begin{array}{l} Given, S_{n} = (pn + {qn}^{2}) \\ Putting n = 1, 2, 3 respectively, we get \\ S_{1} = (p \times 1 + q \times 1^{2}) \\ = p + q \\ S_{2} = (p \times 2 + q \times 2^{2}) \\ = 2 p + 4 q \\ S_{3} = (p \times 3 + q \times 3^{2}) \\ = 3 p + 9 q \\ T_{1} = S_{2} - S_{1} \\ = (2 p + 4 q) - (p + q) \end{array}$ $\begin{array}{l} = p + 3 q \\ T_{2} = S_{3} - S_{2} \\ = (3 p + 9 q) - (2 p + 4 q) \\ = p + 5 q \\ Common difference, \\ d = T_{2} - T_{1} \\ = (p + 5 q) - (p + 3 q) \\ = 2 q \\ Thus, common difference is 2q. \end{array}$

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

Ans

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

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

$\begin{array}{l} first term of A.P. = a \\ and common difference of A.P. = d \\ S_{p} = S_{q} (Given) \\ \Rightarrow \frac{p}{2} {2 a + (p - 1) d} = \frac{q}{2} {2 a + (q - 1) d} \\ \Rightarrow 2 ap + p (p - 1) d = 2 aq + q (q - 1) d \\ \Rightarrow 2 ap - 2 aq = q (q - 1) d - p (p - 1) d \\ \Rightarrow 2 a (p - q) = (q^{2} - q - p^{2} + p) d \\ \Rightarrow 2 a (p - q) = {p - q - (p^{2} - q^{2}) d} \\ \Rightarrow 2 a (p - q) = {(p - q) - (p - q) (p + q) d} \\ \Rightarrow 2 a (p - q) = (p - q) {1 - (p + q) d} \\ \Rightarrow 2 a = {1 - (p + q) d} \end{array}$ $\begin{array}{l} Now, S_{p + q} = \frac{p + q}{2} {2 a + (p + q - 1) d} \\ S_{p + q} = \frac{p + q}{2} [{1 - (p + q)} d + (p + q - 1) d] \\ S_{p + q} = \frac{p + q}{2} {1 - (p + q) + (p + q) - 1} d \\ \Rightarrow S_{p + q} = 0 \\ Thus, the sum of (p + q) terms is zero. \end{array}$

Q.20

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

Ans

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

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

Ans

$\begin{array}{l} Let first term of an A.P. = a \\ and commong difference of A.P. = d \\ It is given that: \\ S_{m} {:S}_{n} = m^{2} : n^{2} \\ \Rightarrow \frac{S_{m}}{S_{n}} = \frac{m^{2}}{n^{2}} \end{array}$ $\begin{array}{l} \Rightarrow \frac{\frac{m}{2} {2 a + (m - 1) d}}{\frac{n}{2} {2 a + (n - 1) d}} = \frac{m^{2}}{n^{2}} \\ \Rightarrow \frac{{2 a + (m - 1) d}}{{2 a + (n - 1) d}} = \frac{m}{n} \\ \Rightarrow 2 an + (mn - n) d = 2 am + (mn - m) d \\ \Rightarrow 2 an - 2 am = (mn - m) d - (mn - n) d \\ \Rightarrow 2 a (n - m) = d (mn - m - mn + n) \\ \Rightarrow 2 a (n - m) = d (- m + n) \\ \Rightarrow 2 a = d \\ Now, \\ \frac{T_{m}}{T_{n}} = \frac{a + (m - 1) 2 a}{a + (n - 1) 2 a} \\ = \frac{a + (m - 1) 2 a}{a + (n - 1) 2 a} \\ = \frac{1 + 2 m - 2}{1 + 2 n - 2} \\ = \frac{2 m - 1}{2 n - 1} \\ Thus, {the ratio of m}^{th} {and n}^{th} term is (2 m - 1) : (2 n - 1) . \end{array}$

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

Ans

$\begin{array}{l} It is given that \\ S_{n} = {3n}^{2} + 5n \\ Replace n by n - 1, we get \\ S_{n-1} = 3 {(n - 1)}^{2} + 5 (n - 1) \\ = {3n}^{2} - 6 n + 3 + 5n - 5 \\ = {3n}^{2} - n - 2 \\ So, T_{n} = S_{n} - S_{n-1} \\ = ({3n}^{2} + 5n) - ({3n}^{2} - n - 2) \\ = 6 n + 2 \\ Since, T_{m} = 164 \\ So, 6m + 2 = 164 \\ \Rightarrow m = \frac{164 - 2}{6} \\ = \frac{162}{6} \\ = 27 \\ Thus, the value of m is 27. \end{array}$

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

Ans

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

Q.24

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

Ans

$\begin{array}{l} A . M . of a and b = \frac{a + b}{2} \\ \frac{a^{n} + b^{n}}{a^{n - 1} + b^{n - 1}} = \frac{a + b}{2} \\ \Rightarrow (a + b) (a^{n - 1} + b^{n - 1}) = 2 (a^{n} + b^{n}) \\ \Rightarrow a^{n} + {ab}^{n - 1} + {ba}^{n - 1} + b^{n} = 2 a^{n} + 2 b^{n} \\ \Rightarrow {ab}^{n - 1} - b^{n} = a^{n} - {ba}^{n - 1} \\ \Rightarrow b^{n - 1} (a - b) = a^{n - 1} (a - b) \\ \Rightarrow b^{n - 1} = a^{n - 1} \\ \Rightarrow {(\frac{a}{b})}^{n - 1} = 1 = {(\frac{a}{b})}^{0} \\ \Rightarrow n = 1 \end{array}$

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 7^th and (m – 1)^th numbers is 5 : 9. Find the value of m.

Ans

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

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

Ans

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

Q.27

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

Ans

$T h e given G .P . is: \frac{5}{2}, \frac{5}{4}, \frac{5}{8}, …$ $\begin{array}{l} a = \frac{5}{2} \\ r = \frac{(\frac{5}{4})}{(\frac{5}{2})} = \frac{1}{2} \\ T_{20} = {ar}^{20 - 1} [âˆµ T_{n} = {ar}^{n - 1}] \\ = \frac{5}{2} {(\frac{1}{2})}^{19} \\ = \frac{5}{2^{20}} \\ and T_{n} = {ar}^{n - 1} \\ = \frac{5}{2} {(\frac{1}{2})}^{n - 1} \\ = \frac{5}{2^{n}} \end{array}$

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

Ans

$\begin{array}{l} Common ratio of given G.P. = 2 \\ 8^{th} term of A.P. = 192 \\ {ar}^{8 - 1} = 192 \\ a . 2^{7} = 192 \\ a = \frac{192}{2^{7}} = 1 .5 \\ 12^{th} term of A.P. = {ar}^{12 - 1} \end{array}$ $\begin{array}{l} = 1 .5 {(2)}^{11} \\ = 1 .5 \times 2048 \\ = 3072 \end{array}$

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

Ans

$\begin{array}{l} Let first term of G.P. be a and common ratio be r. \\ 5^{th} term of G.P. = p \\ \Rightarrow T_{5} = p \\ \Rightarrow p = {ar}^{4} \\ 8^{th} term of G.P. = q \\ \Rightarrow T_{8} = q \\ \Rightarrow q = {ar}^{7} \\ 11^{th} term of G.P. = s \\ \Rightarrow T_{11} = s \\ \Rightarrow s = {ar}^{10} \\ L.H.S. = q^{2} \\ = {({ar}^{7})}^{2} \\ = a^{2} r^{14} \\ R.H.S.= ps \\ = ({ar}^{4}) ({ar}^{10}) \\ = a^{2} r^{14} \\ So, L.H.S.=R.H.S. \\ Hence proved. \end{array}$

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

Ans

$\begin{array}{l} First term of G.P. = - 3 \\ Let common ratio of G.P. = r \\ According to given condition, \\ T_{4} = {(T_{2})}^{2} \\ \Rightarrow {ar}^{3} = {(ar)}^{2} \\ \Rightarrow (- 3) r^{3} = {(- 3)}^{2} r^{2} \\ \Rightarrow r = - 3 \\ Then, T_{7} = {ar}^{6} \\ = (- 3) {(- 3)}^{6} \\ = {(- 3)}^{7} \\ = - 2187 \\ Thus, 7^{th} term of G.P. is - 2187 . \end{array}$

Q.31

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

Ans

$\begin{array}{l} (a) Let t_{n} = 128, a = 2, \\ r = \frac{2 \sqrt{2}}{2} = \sqrt{2} \end{array}$ $\begin{array}{l} t_{n} = {ar}^{n - 1} \\ 128 = 2 {(\sqrt{2})}^{n - 1} \\ \Rightarrow 2^{7} = {(2)}^{1 + \frac{n - 1}{2}} \\ \Rightarrow 7 = 1 + \frac{n - 1}{2} \\ \Rightarrow 6 \times 2 = n - 1 \Rightarrow n = 13 \\ Thus, {128 is 13}^{th} term of G.P. \\ (b) Let t_{n} = 729, a = \sqrt{3}, \\ r = \frac{3}{\sqrt{3}} = \sqrt{3} \\ t_{n} = {ar}^{n - 1} \\ 729 = \sqrt{3} {(\sqrt{3})}^{n - 1} \\ \Rightarrow 3^{6} = {(3)}^{\frac{1}{2} + \frac{n - 1}{2}} \\ \Rightarrow 6 = \frac{1}{2} + \frac{n - 1}{2} \\ \Rightarrow 6 \times 2 = 1 + n - 1 \\ \Rightarrow n = 12 \\ Thus, 729 {is 12}^{th} term of G.P. \\ (c) Let t_{n} = 729, a = \frac{1}{3}, \\ r = \frac{(\frac{1}{9})}{(\frac{1}{3})} = \frac{1}{3} \end{array}$ $\begin{array}{l} \end{array}$ $\begin{array}{l} t_{n} = {ar}^{n - 1} \\ \frac{1}{19683} = \frac{1}{3} {(\frac{1}{3})}^{n - 1} \\ \Rightarrow {(\frac{1}{3})}^{9} = {(\frac{1}{3})}^{n} \\ \Rightarrow 9 = n \\ \Rightarrow n = 9 \\ Thus, \frac{1}{19683} {is 9}^{th} term of G.P. \end{array}$

Q.32

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

Ans

$\begin{array}{l} - \frac{2}{7},x, - \frac{7}{2} are in G.P., then \\ Common ratio = \frac{x}{- (\frac{2}{7})} \\ = - \frac{7 x}{2} \\ Also, common ratio = \frac{(- \frac{7}{2})}{x} \\ = - \frac{7}{2 x} \\ Then, \\ - \frac{7 x}{2} = - \frac{7}{2 x} \\ \Rightarrow x^{2} = 1 \\ \Rightarrow x = \pm 1 \end{array}$

Q.33

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

Ans

$\begin{array}{l} i . \\ 0.15, 0 .0 15, 0 .00 15, \dots 20 terms \\ Since, S_{n} = \frac{a (1 - r^{n})}{1 - r} \\ a = 0 .15, \\ r = \frac{0 .015}{0 .15} = 0 .1 \\ S_{n} = \frac{0 .15 {1 - {(0 .1)}^{20}}}{1 - 0 .1} \\ = \frac{0 .15 {1 - {(0 .1)}^{20}}}{0 .9} \\ = \frac{15}{90} {1 - {(0 .1)}^{20}} \\ = \frac{1}{6} {1 - {(0 .1)}^{20}} \end{array}$ $\begin{array}{l} i i . \\ \sqrt{7}, \sqrt{21}, 3 \sqrt{7}, . .. n terms \\ Since, S_{n} = \frac{a (r^{n} - 1)}{r - 1} \end{array}$ $\begin{array}{l} \end{array}$ $\begin{array}{l} a = \sqrt{7}, \\ r = \frac{\sqrt{21}}{\sqrt{7}} = \sqrt{3} \\ S_{n} = \frac{\sqrt{7} {{(\sqrt{3})}^{n} - 1}}{\sqrt{3} - 1} \times \frac{\sqrt{3} + 1}{\sqrt{3} + 1} \\ = \frac{\sqrt{7} {{(\sqrt{3})}^{n} - 1}}{3 - 1} \times (\sqrt{3} + 1) \end{array}$ $\begin{array}{l} \end{array}$ $\begin{array}{l} iii . \\ 1, - a, a^{2}, - a^{3}, . .. nterms \\ First term of G.P. = 1 \\ Common ratio of G.P. = \frac{- a}{1} \\ = - a \\ Since, S_{n} = \frac{a (1 - r^{n})}{1 - r} \\ S_{n} = \frac{1 {1 - {(- a)}^{n}}}{1 + a} \end{array}$ $\begin{array}{l} \end{array}$ $\begin{array}{l} iv . \\ x^{3}, x^{5}, x^{7}, . .. n terms \\ First term of G.P. = x^{3} \end{array}$ $\begin{array}{l} \end{array}$ $\begin{array}{l} Common ratio of G.P. = \frac{x^{5}}{x^{3}} \\ = x^{2} \\ Since, S_{n} = \frac{a (r^{n} - 1)}{r - 1} \\ S_{n} = \frac{x^{3} {1 - {(x^{2})}^{n}}}{1 - x^{2}} \\ S_{n} = \frac{x^{3} (1 - x^{2 n})}{1 - x^{2}} \end{array}$

Q.34

$Evaluate : \sum_{k = 1}^{11} (2 + 3^{k}) .$

Ans

$\begin{array}{l} We have, \\ \sum_{k = 1}^{11} (2 + 3^{k}) = (2 + 3^{1}) + (2 + 3^{2}) + (2 + 3^{3}) + . .. + (2 + 3^{11}) \\ = 2 \times 11 + (3^{1} + 3^{2} + 3^{3} + . .. + 3^{11}) \\ = 22 + \frac{3 (3^{11} - 1)}{3 - 1} \\ = 22 + \frac{3}{2} (3^{11} - 1) \end{array}$

Q.35

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

Ans

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

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

Ans

$\begin{array}{l} G . P . is 3, 3^{2}, 3^{3}, \dots \\ a = 3, r = \frac{3^{2}}{3} = 3 {and S}_{n} = 120 \\ S_{n} = a \frac{(r^{n} - 1)}{(r - 1)} \end{array}$ $\begin{array}{l} 120 = 3 (\frac{3^{n} - 1}{3 - 1}) \\ 120 = \frac{3}{2} (3^{n} - 1) \\ 120 \times \frac{2}{3} = (3^{n} - 1) \\ 81 = 3^{n} \\ \Rightarrow 3^{4} = 3^{n} \\ \Rightarrow n = 4 \\ Thus, 4 terms of G.P. are needed to give the sum 120. \end{array}$

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

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

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

Ans

$\begin{array}{l} First term of G.P. = (a) \\ = 729 \\ 7^{th} term of G.P. = T_{7} \\ = 64 \\ Then, {ar}^{6} = 64 \\ \Rightarrow 729 r^{6} = 64 \\ r^{6} = \frac{64}{729} \\ r^{6} = \frac{2^{6}}{3^{6}} \\ \Rightarrow r = \frac{2}{3} \\ S_{n} = \frac{a (1 - r^{n})}{(1 - r)} \\ \Rightarrow S_{7} = \frac{729 (1 - \frac{2^{n}}{3^{n}})}{(1 - \frac{2}{3})} \end{array}$ $\begin{array}{l} \Rightarrow S_{7} = 729 \times 3 (1 - \frac{2^{7}}{3^{7}}) \\ \Rightarrow S_{7} = 729 \times 3 (1 - \frac{128}{2187}) \\ \Rightarrow S_{7} = 2187 (\frac{2187 - 128}{2187}) \\ \Rightarrow S_{7} = 2059 \\ Thus, sum of 7 terms of G.P. is 2059. \end{array}$

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

Ans

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

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

Ans

$Let first term of G.P. be ‘a’ and common ratio be ‘r’.$ $\begin{array}{l} T_{4} = x \\ \Rightarrow {ar}^{3} = x \\ T_{10} = y \\ \Rightarrow {ar}^{9} = y \\ T_{16} = z \\ \Rightarrow {ar}^{15} = z \\ Now, \frac{y}{x} = \frac{{ar}^{9}}{{ar}^{3}} \\ = r^{6} \\ and \frac{z}{y} = \frac{{ar}^{15}}{{ar}^{9}} \\ = r^{6} \\ \Rightarrow \frac{y}{x} = \frac{z}{y} \\ \Rightarrow x, y, z are in G.P. \end{array}$

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

Ans

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

Q.42

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

Ans

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

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

Ans

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

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 4^th by 18.

Ans

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

Q.45

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

Ans

$\begin{array}{l} Let first term of G.P. is A and common ratio is R. \\ Given that: \\ T_{p} = a \Rightarrow {AR}^{p - 1} = a \\ T_{q} = b \Rightarrow {AR}^{q - 1} = b \\ T_{r} = c \Rightarrow {AR}^{r - 1} = c \\ L . H . S . = a^{q - r} b^{r - p} c^{p - q} \\ = {({AR}^{p - 1})}^{q - r} {({AR}^{q - 1})}^{r - p} {({AR}^{r - 1})}^{p - q} \\ = A^{q - r + r - p + p - q} R^{(p - 1) (q - r) + (q - 1) (r - p) (r - 1) (p - q)} \\ = A^{0} R^{pq - pr - q + r + qr - qp - r + p + rp - rq - p + q} \\ = A^{0} R^{0} \\ = 1 \times 1 \\ = 1 = R . H . S . \end{array}$

Q.46 If the first and the n^th 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

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

Q.47

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

Ans

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

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

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

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

Ans

$\begin{array}{l} Let two numbers between 3 and 81 are x and y in such a way \\ that 3, x, y, 81 are in G.P. \\ So, first term of G.P. = 3 \\ Common difference = r \\ 4^{th} term of G.P. = {ar}^{3} \\ \Rightarrow 3 r^{3} = 81 \\ \Rightarrow r^{3} = \frac{81}{3} \\ \Rightarrow r = \sqrt[3]{27} \\ = 3 \end{array}$ $\begin{array}{l} ∴ x = T_{2} \\ = ar \\ = 3 \times 3 \\ = 9 \\ andy = T_{3} \\ = {ar}^{2} \\ = 3 \times 3^{2} \\ = 27 \\ Thus, two numbers between 3 and 81 are 9 and 27. \end{array}$

Q.50

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

Ans

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

Q.51

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

Ans

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

Q.52

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

Ans

$\begin{array}{l} Let two numbers are a and b. \\ Then, A.M. = \frac{a + b}{2} \\ \Rightarrow A = \frac{a + b}{2} \\ \Rightarrow a + b = 2 A . .. (i) \\ âˆµ G . M . = \sqrt{ab} \\ \Rightarrow G = \sqrt{ab} \\ \Rightarrow G^{2} = ab . .. (ii) \\ âˆµ a - b = \sqrt{{(a + b)}^{2} - 4 ab} \\ = \sqrt{{(2 A)}^{2} - 4 G^{2}} \\ a - b = 2 \sqrt{A^{2} - G^{2}} . .. (iii) \\ Solving equation (i) and equation (iii), we get \\ a = A + \sqrt{A^{2} - G^{2}} and b = A - \sqrt{A^{2} - G^{2}} \\ Therefore, the two numbers are: \\ A \pm \sqrt{(A + G) (A - G)} . \end{array}$

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 2^nd hour, 4^th hour and n^th hour?

Ans

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

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

Ans

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

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

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

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

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

$\frac{1}{1 \times 2} + \frac{1}{2 \times 3} + \frac{1}{3 \times 4} + …$

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

Ans

$\begin{array}{l} 1 . Let S_{n} = 1 \times 2 + 2 \times 3 + 3 \times 4 + 4 \times 5 + . .. + n \times (n + 1) \\ a_{n} = n \times (n + 1) \\ S_{n} = \sum n \times (n + 1) \\ = \sum n^{2} + \sum n \\ = \frac{n (n + 1) (2 n + 1)}{6} + \frac{n (n + 1)}{2} \\ = \frac{n (n + 1)}{2} (\frac{2 n + 1}{3} + 1) \\ = \frac{n (n + 1)}{2} (\frac{2 n + 4}{3}) \\ = \frac{n (n + 1)}{2} \times 2 (\frac{n + 2}{3}) \\ S_{n} = \frac{1}{3} n (n + 1) (n + 2) \\ 2. Let S_{n} = 1 \times 2 \times 3 + 2 \times 3 \times 4 + 3 \times 4 \times 5 + . .. + n \times (n + 1) \times (n + 2) \\ a_{n} = n \times (n + 1) \times (n + 2) \\ S_{n} = \sum n \times (n + 1) \times (n + 2) \\ = \sum (n^{3} + 3 n^{2} + 2 n) \\ = \sum n^{3} + 3 \sum n^{2} + 2 \sum n \\ = {\{\frac{n (n + 1)}{2}\}}^{2} + 3 . \frac{n (n + 1) (2 n + 1)}{6} + 2 . \frac{n (n + 1)}{2} \\ = \frac{n^{2} {(n + 1)}^{2}}{4} + \frac{n (n + 1) (2 n + 1)}{2} + \frac{2 n (n + 1)}{2} \\ = \frac{n (n + 1)}{2} \{\frac{n (n + 1)}{2} + (2 n + 1) + 2\} \\ = \frac{n (n + 1)}{2} \{\frac{n^{2} + n + 4 n + 2 + 4}{2}\} \\ = \frac{n (n + 1)}{2} (\frac{n^{2} + 5 n + 6}{2}) \\ = \frac{n (n + 1) (n + 2) (n + 3)}{4} \\ S_{n} = \frac{1}{4} n (n + 1) (n + 2) (n + 3) \\ 3 . a_{n} = (2 n + 1) \times n^{2} \\ S_{n} = \sum (2 n^{3} + n^{2}) \\ = 2 \sum n^{3} + \sum n^{2} \\ = 2 {\{\frac{n (n + 1)}{2}\}}^{2} + \frac{n (n + 1) (2 n + 1)}{6} \\ = 2 . \frac{n^{2} {(n + 1)}^{2}}{4} + \frac{n (n + 1) (2 n + 1)}{6} \\ = \frac{n^{2} {(n + 1)}^{2}}{2} + \frac{n (n + 1) (2 n + 1)}{6} \\ = \frac{n (n + 1)}{2} \{n (n + 1) + \frac{(2 n + 1)}{3}\} \\ = \frac{n (n + 1)}{2} \{\frac{3 n (n + 1) + (2 n + 1)}{3}\} \\ = \frac{n (n + 1)}{2} \{\frac{3 n^{2} + 3 n + 2 n + 1}{3}\} \\ = \frac{n (n + 1)}{2} \{\frac{3 n^{2} + 5 n + 1}{3}\} \\ S_{n} = \frac{1}{6} n (n + 1) (3 n^{2} + 5 n + 1) \\ 4 . Let S_{n} = \frac{1}{1 \times 2} + \frac{1}{2 \times 3} + \frac{1}{3 \times 4} + . .. + \frac{1}{n (n + 1)} \\ = (\frac{1}{1} - \frac{1}{2}) + (\frac{1}{2} - \frac{1}{3}) + (\frac{1}{3} - \frac{1}{4}) + . .. + (\frac{1}{n} - \frac{1}{n + 1}) \\ = \frac{1}{1} - \frac{1}{n + 1} \\ = \frac{n + 1 - 1}{n + 1} \\ = \frac{n}{n + 1} \\ 5. 5^{2} + 6^{2} + 7^{2} + . .. + 20^{2} \\ = 1^{2} + 2^{2} + 3^{2} + 4^{2} + 5^{2} + 6^{2} + 7^{2} + . .. + 20^{2} - (1^{2} + 2^{2} + 3^{2} + 4^{2}) \\ = \sum 20^{2} - \sum 4^{2} \\ = \frac{20 (20 + 1) (2 \times 20 + 1)}{6} - \frac{4 (4 + 1) (2 \times 4 + 1)}{6} \\ = \frac{20 \times 21 \times 41}{6} - \frac{4 \times 5 \times 9}{6} \\ = \frac{17220 - 180}{6} \\ = \frac{17040}{6} \\ = 2840 \\ 6 . {Let S}_{n} = 3 \times 8 + 6 \times 11 + 9 \times 14 + . .. \\ T_{n} = (3, 6, 9, . .. n^{th} term) (8, 11, 14, . .. n^{th} term) \\ = \{3 + (n - 1) 3\} \{8 + (n - 1) 3\} \\ = (3 n) (3 n + 5) \\ = 9 n^{2} + 15 n \\ S_{n} = \sum T_{n} \\ = \sum (9 n^{2} + 15 n) \\ = 9 \sum n^{2} + 15 \sum n \\ = 9 \times \frac{n (n + 1) (2 n + 1)}{6} + 15 \times \frac{n (n + 1)}{2} \\ = 3 \times \frac{n (n + 1) (2 n + 1)}{2} + 15 \times \frac{n (n + 1)}{2} \\ = \frac{3 n (n + 1)}{2} \{(2 n + 1) + 5\} \\ = \frac{3 n (n + 1)}{2} (2 n + 6) \\ = 3 n (n + 1) (n + 3) \\ 7 . Let S_{n} = 1^{2} + (1^{2} + 2^{2}) + (1^{2} + 2^{2} + 3^{2}) + . .. + (1^{2} + 2^{2} + 3^{2} + . .. + n^{2}) \\ Here, T_{n} = 1^{2} + 2^{2} + 3^{2} + . .. + n^{2} \\ = \sum n^{2} \\ = \frac{n (n + 1) (2 n + 1)}{6} \\ = \frac{1}{6} n (2 n^{2} + 3 n + 1) \\ = \frac{1}{6} (2 n^{3} + 3 n^{2} + n) \\ S_{n} = \sum T_{n} \\ = \frac{1}{6} \sum (2 n^{3} + 3 n^{2} + n) \\ = \frac{1}{6} (2 \sum n^{3} + 3 \sum n^{2} + \sum n) \\ = \frac{1}{6} [2 {\{\frac{n (n + 1)}{2}\}}^{2} + 3 \frac{n (n + 1) (2 n + 1)}{6} + \frac{n (n + 1)}{2}] \\ = \frac{1}{6} \{\frac{n^{2} {(n + 1)}^{2}}{2} + \frac{n (n + 1) (2 n + 1)}{2} + \frac{n (n + 1)}{2}\} \\ = \frac{1}{6} \times \frac{n (n + 1)}{2} \{n (n + 1) + (2 n + 1) + 1\} \\ = \frac{1}{6} \times \frac{n (n + 1)}{2} (n^{2} + n + 2 n + 2) \\ = \frac{1}{12} n (n + 1) (n^{2} + 3 n + 2) \\ = \frac{1}{12} n (n + 1) (n + 1) (n + 2) \\ S_{n} = \frac{n {(n + 1)}^{2} (n + 2)}{12} \end{array}$

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

1. n (n + 1) (n + 4).
2. n² + 2ⁿ
^3.^{(2n – 1)²}

Ans

${Let T}_{n} = n (n + 1) (n + 4)$ $\begin{array}{l} = n (n^{2} + 5 n + 4) \\ = n^{3} + 5 n^{2} + 4 n \\ S_{n} = \sum T_{n} \\ = \sum (n^{3} + 5 n^{2} + 4 n) \\ = \sum n^{3} + 5 \sum n^{2} + 4 \sum n \\ = {\frac{n (n + 1)}{2}}^{2} + \frac{5 n (n + 1) (2 n + 1)}{6} + \frac{4 n (n + 1)}{2} \\ = \frac{n (n + 1)}{2} {\frac{n (n + 1)}{2} + 5 . \frac{(2 n + 1)}{3} + 4} \\ = \frac{n (n + 1)}{2} {\frac{3 n (n + 1) + 10 (2 n + 1) + 24}{6}} \\ = \frac{n (n + 1)}{2} (\frac{3 n^{2} + 3 n + 20 n + 10 + 24}{6}) \\ S_{n} = \frac{n (n + 1)}{12} (3 n^{2} + 23 n + 34) \end{array}$

$\begin{array}{l} Let T_{n} = n^{2} + 2^{n} \\ S_{n} = \sum T_{n} \\ = \sum (n^{2} + 2^{n}) \\ = \sum n^{2} + \sum 2^{n} \\ = \frac{n (n + 1) (2 n + 1)}{6} + (2 + 2^{2} + 2^{3} + 2^{4} + . .. + 2^{n}) \end{array}$ $\begin{array}{l} \end{array}$ $\begin{array}{l} = \frac{n (n + 1) (2 n + 1)}{6} + \frac{2 (2^{n} - 1)}{2 - 1} [âˆµ S_{n} = \frac{a (r^{n} - 1)}{r - 1}] \\ = \frac{n (n + 1) (2 n + 1)}{6} + 2 (2^{n} - 1) \\ ∴ S_{n} = \frac{1}{6} n (n + 1) (2 n + 1) + 2 (2^{n} - 1) \end{array}$

$\begin{array}{l} Let t_{n} = {(2n - 1)}^{2} \\ = 4 n^{2} - 4 n + 1 \\ S_{n} = \sum (4 n^{2} - 4 n + 1) \\ = 4 \sum n^{2} - 4 \sum n + \sum 1 \\ = 4 \frac{n (n + 1) (2 n + 1)}{6} - 4 \frac{n (n + 1)}{2} + n \\ = 2 \frac{n (n + 1) (2 n + 1)}{3} - 2 n (n + 1) + n \\ = \frac{2 n (2 n^{2} + 3 n + 1)}{3} - 2 n^{2} - 2 n + n \\ = \frac{2 (2 n^{3} + 3 n^{2} + n)}{3} - 2 n^{2} - n \\ = \frac{4 n^{3} + 6 n^{2} + 2 n - 6 n^{2} - 3 n}{3} \\ = \frac{4 n^{3} - n}{3} \end{array}$ $\begin{array}{l} = \frac{n}{3} (4 n^{2} - 1) \\ S_{n} = \frac{n}{3} (2 n - 1) (2 n + 1) \end{array}$

Q.58

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

Ans

$\begin{array}{l} 1 . Given G.P. is: \\ 1, \frac{1}{3}, \frac{1}{9}, . .. \\ a = 1, r = \frac{(\frac{1}{3})}{1} = \frac{1}{3} \\ S_{\infty} = \frac{a}{1 - r} \\ = \frac{1}{1 - \frac{1}{3}} \\ = \frac{1}{\frac{2}{3}} \\ = \frac{3}{2} = 1 .5 \\ 2. The given G.P. is: \\ 6, 1.2, 0.24, \dots \\ a = 6, \\ r = \frac{1 .2}{6} = 0 .2 \\ S_{\infty} = \frac{a}{1 - r} \\ = \frac{6}{1 - 0 .2} \\ = \frac{6}{0 .8} \\ = \frac{60}{8} \\ = 7 .5 \\ Thus, the sum of infinite terms is 7.5. \\ 3. The given G.P. is: \\ 5, \frac{20}{7}, \frac{80}{49}, . .. \\ a = 5, \\ r = \frac{(\frac{20}{7})}{5} = \frac{4}{7} \\ S_{\infty} = \frac{a}{1 - r} \\ = \frac{5}{1 - \frac{4}{7}} \\ = \frac{5}{(\frac{3}{7})} \\ = \frac{35}{3} \\ = 11 .6 \\ Thus, the sum of infinite terms is 11.6. \\ 4. The given G.P. is: \\ \frac{- 3}{4}, \frac{3}{16}, \frac{- 3}{64}, . .. \\ a = \frac{- 3}{4}, \\ r = \frac{(\frac{3}{16})}{(\frac{- 3}{4})} \\ = - \frac{1}{4} \\ S_{\infty} = \frac{a}{1 - r} = \frac{(- \frac{3}{4})}{1 - (\frac{- 1}{4})} = - \frac{3}{4} \times \frac{4}{5} = - \frac{3}{5} \\ 5. L.H.S.= 3^{\frac{1}{2}} \times 3^{\frac{1}{4}} \times 3^{\frac{1}{8}} . .. \\ = 3^{\frac{1}{2} + \frac{1}{4} + \frac{1}{8} + . ..} \\ = 3^{(\frac{\frac{1}{2}}{1 - \frac{1}{2}})} [âˆµ S \infty = \frac{a}{1 - r}] \\ = 3^{(\frac{\frac{1}{2}}{\frac{1}{2}})} \\ = 3^{1} \\ = 3 = R . H . S . \\ Hence, it was to be proved. \\ 6. x = 1 + a + a^{2} + . .. \\ = \frac{1}{1 - a} [S_{\infty} = \frac{a}{1 - r}] \\ and y = 1 + b + b^{2} + . .. \\ = \frac{1}{1 - b} \\ L . H . S . = 1 + ab + a^{2} b^{2} + . .. \\ = \frac{1}{1 - ab} \\ R . H . S . = \frac{xy}{x + y - 1} \\ = \frac{(\frac{1}{1 - a}) (\frac{1}{1 - b})}{(\frac{1}{1 - a}) + (\frac{1}{1 - b}) - 1} \\ = \frac{1}{1 - b + 1 - a - (1 - a) (1 - b)} \\ = \frac{1}{2 - b - a - (1 - b - a + ab)} \\ = \frac{1}{2 - b - a - 1 + b + a - ab} \\ = \frac{1}{1 - ab} \\ So,L.H.S.=R.H.S. \end{array}$

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

Ans

$\begin{array}{l} Let first term of A.P. be a and common difference be d. \\ Then, \\ T_{m+n} + T_{m-n} = {a + (m + n - 1) d} + {a + (m - n - 1) d} \\ = 2 a + (m + n - 1 + m - n - 1) d \\ = 2 a + 2 (m - 1) d \\ = 2 {a + (m - 1) d} \\ = 2 T_{m} \\ Thus, \\ T_{m+n} + T_{m-n} = 2 T_{m} \\ Hence proved. \end{array}$

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

Ans

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

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

Ans

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

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

Ans

$\begin{array}{l} The numbers between 200 to 400 divisible by 7 are: \\ 203, 210, 217, 224, 231, . .., 399 . \\ Here, a = 203, d = 210 - 203 = 7, l = 399 \\ âˆµ l = a + (n - 1) d \\ 399 = 203 + (n - 1) 7 \\ \Rightarrow \frac{196}{7} = n - 1 \end{array}$ $\begin{array}{l} \Rightarrow 28 + 1 = n \\ \Rightarrow n = 29 \\ S_{n} = \frac{n}{2} (a + l) \\ \Rightarrow S_{29} = \frac{29}{2} (203 + 399) \\ = \frac{29}{2} \times 602 \\ = 8729 \\ Thus, the sum of multiples of 7 between 200 \\ and 400 is 8729. \end{array}$

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

Ans

$\begin{array}{l} The numbers from 1 to 100 divisible by 2: \\ 2, 4, 6, 8, . .., 100 \\ Here, a = 2, d = 4 - 2 = 2, l = 100 \\ âˆµ l = a + (n - 1) d \\ 100 = 2 + (n - 1) 2 \\ \frac{98}{2} = n - 1 \\ \Rightarrow n = 49 + 1 \\ = 50 \\ âˆµ S_{n} = \frac{n}{2} (a + l) \end{array}$ $\begin{array}{l} ∴ S_{50} = \frac{50}{2} (2 + 100) \\ = 25 \times 102 \\ = 2550 \\ The numbers from 1 to 100 divisible by 5: \\ 5, 10, 15, 20, . .., 100 \\ Here, a = 5, d = 10 - 5 = 5, l = 100 \\ âˆµ l = a + (n - 1) d \\ 100 = 5 + (n - 1) 5 \\ \frac{95}{5} = n - 1 \\ \Rightarrow n = 19 + 1 \\ = 20 \\ âˆµ S_{n} = \frac{n}{2} (a + l) \\ ∴ S_{20} = \frac{20}{2} (5 + 100) \\ = 10 \times 105 \\ = 1050 \\ The numbers from 1 to 100 divisible by 2 and 5: \\ 10,20,30, . ..,100 \\ Here, a = 10, d = 20 - 10 = 10, l = 100 \\ âˆµ l = a + (n - 1) d \\ 100 = 10 + (n - 1) 10 \\ \Rightarrow \frac{90}{10} = n - 1 \end{array}$ $\begin{array}{l} \end{array}$ $\begin{array}{l} \Rightarrow n = 10 \\ âˆµ S_{n} = \frac{n}{2} (a + l) \\ ∴ S_{10} = \frac{10}{2} (10 + 100) \\ = 5 \times 110 \\ = 550 \\ Sum of numbers divisible by 2 or 5 \\ = Sum of multiples of 2 + Sum of multiples of 5 \\ - Sum of multiples of 10 \\ = 2550 + 1050 - 550 \\ = 3050 \end{array}$

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

Ans

$\begin{array}{l} Two digit numbers which when divided by 4, yields \\ 1 as remainder are: \\ 13, 17, 21, . .., 97 \\ a = 13, d = 17 - 13 = 4, l = 97 \\ âˆµ l = a + (n - 1) d \\ 97 = 13 + (n - 1) d \\ 97 - 13 = (n - 1) 4 \\ \frac{84}{4} = n - 1 \\ n = 21 + 1 \\ = 22 \end{array}$ $\begin{array}{l} âˆµ S_{n} = \frac{n}{2} (a + l) \\ ∴ S_{22} = \frac{22}{2} (13 + 97) \\ = \frac{22}{2} \times 110 \\ = 1210 \\ Thus, the sum of 22 terms is 1210. \end{array}$

Q.65

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

Ans

$\begin{array}{l} Since, f (x + y) = f (x) f (y), f (1) = 3 and \sum_{x = 1}^{n} f (x) = 120 . \\ So, f (2) = f (1 + 1) \\ = f (1) f (1) \\ = 3 \times 3 \\ = 3^{2} \\ f (3) = f (2 + 1) \\ = f (2) f (1) \\ = 3^{2} \times 3 \\ = 3^{3} \\ f (4) = f (2 + 2) \\ = f (2) f (2) \end{array}$ $\begin{array}{l} = 3^{2} \times 3^{2} \\ = 3^{4} \\ âˆµ \sum_{x = 1}^{n} f (x) = 120 \\ \Rightarrow f (1) + f (2) + f (3) + f (4) + . .. + f (n) = 120 \\ 3 + 3^{2} + 3^{3} + 3^{4} + . .. + 3^{n} = 120 \end{array}$ $\begin{array}{l} \end{array}$ $\begin{array}{l} \Rightarrow \frac{3 (3^{n} - 1)}{3 - 1} = 120 \\ \Rightarrow 3^{n} - 1 = 120 \times \frac{2}{3} \\ \Rightarrow 3^{n} = 80 + 1 \\ \Rightarrow 3^{n} = 3^{4} \\ \Rightarrow n = 4 \\ Thus, the value of n is 4. \end{array}$

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

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

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

$\begin{array}{l} First term of G.P. = 1 \\ Let common ratio = r \\ T_{3} + T_{5} = 90 \\ \Rightarrow {ar}^{2} + {ar}^{4} = 90 \\ \Rightarrow 1 . r^{2} + 1 . r^{4} = 90 \\ \Rightarrow r^{4} + r^{2} = 90 \\ \Rightarrow r^{4} + r^{2} - 90 = 0 \\ \Rightarrow (r^{2} + 10) (r^{2} - 9) = 0 \\ \Rightarrow r^{2} = 9, - 10 \end{array}$ $\begin{array}{l} \Rightarrow r = \pm 3 [Neglecting r^{2} = - 10] \\ Thus, the common ratio of G.P. is \pm 3 . \end{array}$

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

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

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

$\begin{array}{l} Let G.P. containing even number of terms is: \\ {a, ar, ar}^{2} {, ar}^{3}, . .. {, ar}^{n}, {ar}^{n+1}, . \dots, {ar}^{2 n - 1}, {ar}^{2 n} \\ According to given condition, \\ S_{2 n} = 5 \times (a + {ar}^{2} + {ar}^{4} + . .. + {ar}^{2 n - 1}) \\ (\begin{array}{l} a + ar + {ar}^{2} + {ar}^{3} + . .. + {ar}^{n} \\ + {ar}^{n+1} + . .. + {ar}^{2 n - 1} + {ar}^{2 n} \end{array}) = 5 \times (a + {ar}^{2} + {ar}^{4} + . .. + {ar}^{2 n - 1}) \end{array}$ $\begin{array}{l} (ar + {ar}^{3} + . .. + {ar}^{n} + . .. + {ar}^{2 n}) = 4 \times (a + {ar}^{2} + {ar}^{4} + . .. + {ar}^{2 n - 1}) \\ \frac{ar (r^{2 n} - 1)}{r - 1} = 4 \times \frac{a {{(r^{2})}^{n} - 1}}{r - 1} \\ \Rightarrow r (r^{2 n} - 1) = 4 (r^{2 n} - 1) \\ \Rightarrow r = 4 \\ Thus, the common ratio of G.P. is 4. \end{array}$

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

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

Q.71

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

Ans

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

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

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

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

Ans

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

Q.74

$If a (\frac{1}{b} + \frac{1}{c}), b (\frac{1}{c} + \frac{1}{a}), c (\frac{1}{a} + \frac{1}{b}) are in A.P., prove that a, b and c are in A.P.$

Ans

$\begin{array}{l} Since,a (\frac{1}{b} + \frac{1}{c}), b (\frac{1}{c} + \frac{1}{a}),c (\frac{1}{a} + \frac{1}{b}) areinA.P. \\ \Rightarrow (\frac{a}{b} + \frac{a}{c}), (\frac{b}{c} + \frac{b}{a}), (\frac{c}{a} + \frac{c}{b}) areinA.P. \\ \Rightarrow (\frac{a}{a} + \frac{a}{b} + \frac{a}{c}), (\frac{b}{b} + \frac{b}{c} + \frac{b}{a}), (\frac{c}{c} + \frac{c}{a} + \frac{c}{b}) areinA.P. \\ \Rightarrow a (\frac{1}{a} + \frac{1}{b} + \frac{1}{c}),b (\frac{1}{b} + \frac{1}{c} + \frac{1}{a}), c (\frac{1}{c} + \frac{1}{a} + \frac{1}{b}) areinA.P. \\ \Rightarrow a,b,careinA.P. \end{array}$

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

Ans

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

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

Ans

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

Q.77

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

Ans

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

Q.78

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

Ans

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

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

Ans

$\begin{array}{l} (i) Let S_{n} = 5 + 55 + 555 + \dots \\ = 5 (1 + 11 + 111 + 1111 + . ..) \\ = \frac{5}{9} (9 + 99 + 999 + 9999 + . ..) \\ = \frac{5}{9} {(10 - 1) + (10^{2} - 1) + (10^{3} - 1) + (10^{4} - 1) + . ..} \\ = \frac{5}{9} {10 + 10^{2} + 10^{3} + 10^{4} + . .. - 1 - 1 - 1 - 1 - . .. n times} \\ = \frac{5}{9} {\frac{10 (10^{n} - 1)}{10 - 1} - n} [âˆµ S_{n} = \frac{a (r^{n} - 1)}{r - 1}] \\ = \frac{5}{9} {\frac{10 (10^{n} - 1)}{9} - n} \\ = \frac{50}{81} (10^{n} - 1) - \frac{5 n}{9} \\ (ii) Let S_{n} = 0.6 + 0.66 + 0.666 + \dots \\ = 6 (0 .1 + 0 .11 + 111 + 1111 + . ..) \\ = \frac{6}{9} (0 .9 + 0 .99 + 0 .999 + 0 .9999 + . ..) \\ = \frac{2}{3} {(1 - 0 .1) + (1 - 0 .01) + (1 - 0 .001) + (1 - 0 .0001) + . ..} \\ = \frac{2}{3} {1 + 1 + 1 + 1 + . .. n times - \frac{1}{10} - \frac{1}{10^{2}} - \frac{1}{10^{3}} - \frac{1}{10^{4}} - . ..} \end{array}$ $\begin{array}{l} = \frac{2}{3} {n - \frac{\frac{1}{10} (1 - \frac{1}{10^{n}})}{1 - \frac{1}{10}}} [âˆµ S_{n} = \frac{a (r^{n} - 1)}{r - 1}] \\ = \frac{2}{3} n - \frac{2}{3} \times \frac{(1 - \frac{1}{10^{n}})}{9} \\ = \frac{2}{3} n - \frac{2}{27} \times (1 - \frac{1}{10^{n}}) \end{array}$

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

Ans

$\begin{array}{l} 2 \times 4 + 4 \times 6 + 6 \times 8 + . .. + n terms \\ T_{n} = {2 + (n - 1) 2} {4 + (n - 1) 2} \\ = (2 n) (2 n + 2) \\ = 4 n (n + 1) \\ Putting n = 20, we get \\ T_{20} = 4 (20) (20 + 1) \\ = 80 \times 21 \\ = 1680 \\ Thus, {the 20}^{th} term of series is 1680. \end{array}$

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

Ans

$\begin{array}{l} Let S_{n} = 3 + 7 + 13 + 21 + 31 + . .. + a_{n - 1} + a_{n} \\ and - \underline{S_{n} = - 3 - 7 - 13 - 21 - 31 - . .. - a_{n - 1} - a_{n}} \\ 0 = 3 + \{4 + 6 + 8 + 10 + . . . (n - 1) terms\} - a_{n} \\ a_{n} = 3 + \frac{n - 1}{2} \{8 + (n - 2) 2\} \\ = 3 + (n - 1) (4 + n - 2) \\ = 3 + (n - 1) (n + 2) \\ a_{n} = 3 + n^{2} + n - 2 \\ = n^{2} + n + 1 \\ S_{n} = \sum n^{2} + \sum n + \sum 1 \\ = \frac{n (n + 1) (2 n + 1)}{6} + \frac{n (n + 1)}{2} + n \\ = \frac{n}{6} \{(n + 1) (2 n + 1) + 3 (n + 1) + 6\} \\ = \frac{n}{6} (2 n^{2} + 3 n + 1 + 3 n + 3 + 6) \\ = \frac{n}{6} (2 n^{2} + 6 n + 10) \\ S_{n} = \frac{n}{3} (n^{2} + 3 n + 5) \end{array}$

Q.82

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

Ans

$\begin{array}{l} We are given that \\ S_{1} = \sum n \\ = \frac{n (n + 1)}{2} \\ S_{2} = \sum n^{2} \\ = \frac{n (n + 1) (2 n + 1)}{6} \\ S_{3} = \sum n^{3} \\ = {\frac{n (n + 2)}{2}}^{2} \\ R . H . S . = S_{3} (1 + 8 S_{1}) \\ = {\frac{n (n + 2)}{2}}^{2} {1 + 8 \times \frac{n (n + 1)}{2}} \\ = \frac{n^{2} {(n + 2)}^{2}}{4} {1 + 4 n (n + 1)} \\ = \frac{n^{2} {(n + 2)}^{2}}{4} (1 + 4 n^{2} + 4 n) \\ = \frac{n^{2} {(n + 2)}^{2}}{4} (4 n^{2} + 4 n + 1) \\ = \frac{n^{2} {(n + 2)}^{2} {(2 n + 1)}^{2}}{4} \\ = {3 \times \frac{n (n + 2) (2 n + 1)}{6}}^{2} \end{array}$ $\begin{array}{l} = 9 {\frac{n (n + 2) (2 n + 1)}{6}}^{2} \\ = 9 S_{2}^{2} = L . H . S . \\ This was to be proved. \end{array}$

Q.83

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

Ans

$\begin{array}{l} \frac{1^{3}}{1} + \frac{1^{3} + 2^{3}}{1 + 3} + \frac{1^{3} + 2^{3} + 3^{3}}{1 + 3 + 5} + . .. \\ T_{n} = \frac{1^{3} + 2^{3} + 3^{3} + . .. + n^{3}}{1 + 3 + 5 + . ..} \\ = \frac{\sum n^{3}}{\frac{n}{2} {2 \times 1 + (n - 1) 2}} \\ = \frac{{\frac{n (n + 1)}{2}}^{2}}{\frac{n}{2} (2 + 2 n - 2)} \\ = \frac{n^{2} {(n + 1)}^{2}}{4 n^{2}} \\ = \frac{1}{4} (n^{2} + 2 n + 1) \end{array}$ $\begin{array}{l} S_{n} = \frac{1}{4} (\sum n^{2} + 2 \sum n + \sum 1) \\ = \frac{1}{4} [\frac{n (n + 1) (2 n + 1)}{6} + \frac{2 n (n + 1)}{2} + n] \\ = \frac{1}{4} {\frac{n (n + 1) (2 n + 1)}{6} + n + 1 + 1} \\ = \frac{n}{4} (\frac{2 n^{2} + 3 n + 1 + 6 n + 12}{6}) \\ = \frac{n}{4} (\frac{2 n^{2} + 9 n + 13}{6}) \\ = \frac{n}{24} (2 n^{2} + 9 n + 13) \end{array}$

Q.84

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

Ans

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

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

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

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NCERT Solutions Class 11 Maths Chapter 9

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

Key Topics Covered In Chapter 9 Class 11 Mathematics

NCERT Solutions for Class 11 Mathematics Chapter 9 Exercise & Solutions

NCERT Exemplar Class 11 Mathematics

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NCERT Solutions Class 11 Maths Chapter 9

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

Key Topics Covered In Chapter 9 Class 11 Mathematics

NCERT Solutions for Class 11 Mathematics Chapter 9 Exercise & Solutions

NCERT Exemplar Class 11 Mathematics

Share

Please register to view this section

FAQs (Frequently Asked Questions)

1. How can I make my NCERT Class 11 Mathematics strong?

2. How will the NCERT Solutions for Class 11 Mathematics Chapter 9 benefit me?

Share

NCERT Solutions Related Links

NCERT Solutions

Solved Board Papers

CBSE Class

ICSE Class

Sample Paper

Exam Weightage

Board Paper Solution 2020

Test Preparation

Other Services

About Us

Terms & Conditions