# NCERT Solutions for Class 11 Maths Chapter 9 Sequences and Series

NCERT solutions for class 11 maths chapter 9 Sequences and Series: Sequence means the progression of numbers in a definite order and series means the sum of the objects of the sequence. In the previous classes, you have studied about arithmetic progression(A.P). In this chapter, we will discuss more arithmetic progression(A.P) and geometric progression(G.P). In this article, you will get NCERT solutions for class 11 maths chapter 9 sequences and series. Important topics like arithmetic progression(A.P), geometric progression(G.P), arithmetic means(A.M), geometric mean(G.M), the relationship between A.M. and G.M, sum to n terms of special series, sum to n terms of squares and cubes of natural numbers are covered in this chapter. You will get questions related to these topics in the solutions of NCERT for class 11 maths chapter 9 sequences and series. In this chapter, there are two types of sequence.

• Finite sequence( A sequence containing a finite number of terms)
• Infinite sequence( A sequence has a first term but doesn't have last term or a sequence which is not finite)

Check all NCERT solutions from class 6 to 12 to understand the concepts in a much easy way. There are four exercises and a miscellaneous exercise in this chapter which are explained below.

Exercise:9.1

Exercise:9.2

Exercise:9.3

Exercise:9.4

Miscellaneous Exercise

## Topics of NCERT Grade 11 Maths Chapter- 9 Sequences and Series

9.1 Introduction

9.2 Sequences

9.3 Series

9.4 Arithmetic Progression (A.P.)

9.5 Geometric Progression (G.P.)

9.6 Relationship Between A.M. and G.M.

9.7 Sum to n terms of Special Series

## Question:1  Write the first five terms of each of the sequences in Exercises 1 to 6 whose nth terms are:

Given :

Therefore, the required number of terms =3, 8, 15, 24, 35

Given :

Therefore, the required number of terms

Given :

Therefore, required number of terms

Given :

Therefore, the required number of terms

Given :

Therefore, the required number of terms

Given :

Therefore, the required number of terms

Put

Put  n=24,

Hence, we have

Given :

Put n=7,

Heence, we have

Given :

Put n =9,

The value of

Given :

Put n=20,

Hence, value of

Given :

Hence, five terms of series are

Series

Given :

Hence, five terms of series are

Series

Given :

Hence, five terms of series are

Series

Question:14 The Fibonacci sequence is defined by

Find, for n = 1, 2, 3, 4, 5

Given : The Fibonacci sequence is defined by

Solutions of NCERT for class 11 maths chapter 9 sequences and series-Exercise: 9.2

Odd integers from 1 to 2001 are

This sequence is an A.P.

Here , first term =a =1

common difference = 2.

We know ,

The , sum of odd integers from 1 to 2001 is 1002001.

Numbers divisible by 5 from 100 to 1000 are

This sequence is an A.P.

Here , first term =a =105

common difference = 5.

We know ,

The  sum of numbers divisible by 5 from 100 to 1000 is 98450.

First term =a=2

Let the series be

Sum of first five terms

Sum of next five terms

Given : The sum of the first five terms is one-fourth of  the next five terms.

To prove  :

L.H.S :

Hence, 20th term is –112.

Given : A.P. =

Given : sum = -25

Given : In an A.P., if pth term is 1/q  and qth term is 1/p

Subtracting (2) from (1), we get

Putting value of d in equation (1),we get

Hence,the sum of first pq  terms is 1/2  (pq +1), where .

Given : A.P. 25, 22, 19, ….....

a=25  , d = -3

n could not be  so n=8.

Last term

The, last term of A.P. is 4.

Given :

Comparing LHS and RHS , we have

and

Putting value of d,

If the sum of n terms of an A.P. is ,

Comparing coefficients of  on both side , we get

The common difference of AP is 2q.

Given: The sums of n terms of two arithmetic progressions are in the ratio.

There are two AP's with first terms =    and common difference =

Substituting n=35,we get

Thus, the ratio of the 18th term of AP's is

Let first term of AP = a and common difference = d.

Then,

Given :

Now,

Thus, sum of p+q terms of AP is 0.

To prove :

Let  and d be the first term and the common difference of AP, respectively.

According to the given information, we have

Subtracting equation (2) from (1), we have

Subtracting equation (3) from (2), we have

Equating values of d, we have

Dividing both sides from pqr, we get

Hence, the given result is proved.

Let a and b be the first term and common difference of a AP ,respectively.

Given : The ratio of the sums of m and n terms of an A.P. is  .

To prove :  the ratio of mth and nth term is .

Put , we get

From equation (1) ,we get

Hence proved.

Given : If the sum of n terms of an A.P. is    and its term is 164

Let a and d be first term and common difference of a AP ,respectively.

Sum of n terms =

Comparing the coefficients of n on both side , we have

Also ,

m th term is 164.

Hence, the value of m is 27.

Let five numbers be A,B,C,D,E.

Then

Here we have,

Thus, we have

Thus, the five numbers are 11,14,17,20,23.

Given :    is the A.M. between a and b.

Thus, value of n is 1.

Let A,B,C.........M be m numbers.

Then,

Here we have,

Given : the ratio of and    numbers is 5 : 9.

Putting value of d from above,

Thus, value of m is 14.

The  first instalment is of Rs. 100.

If the instalment increase by Rs 5 every month, second instalment is Rs.105.

Then , it forms an AP.

We have ,

Thus, he will pay  Rs. 245 in the 30th instalment.

The angles of polygon forms AP with common difference of  and first term as  .

We know that sum of angles of polygon with n sides is

Sides of polygon are 9 or 16.

CBSE NCERT solutions for class 11 maths chapter 9 sequences and series-Exercise: 9.3

Question:1 Find the and terms of the G.P.

G.P :

first term = a

common ratio =r

the nth term of G.P

First term = a

common ratio =r=2

term is 192

is the  term of a G.P.

To prove :

Let first term=a and common ratio = r

Dividing equation 2 by 1, we have

Dividing equation 3 by 2, we have

Equating values of  ,  we have

Hence proved

First term =a= -3

term of a G.P. is square of its second term

Thus, seventh term is -2187.

Question:5(a) Which term of the following sequences:

Given :

n th term is given as 128.

The, 13 th term is 128.

Question:5(b) Which term of the following sequences:

Given :

n th term is given as 729.

The, 12 th term is 729.

Question:5(c) Which term of the following sequences:

Given :

n th term is given as

Thus, n=9.

Common ratio=r.

Thus, for  ,given numbers will be in GP.

geometric progressions is  0.15, 0.015, 0.0015, ... .....

a=0.15 , r = 0.1 , n=20

The sum to the indicated number of terms in each of the geometric progressions is:

Question:11 Evaluate

Given :

These terms form GP with a=3 and r=3.

Given : The sum of first three terms of a G.P. is and their product is 1.

Let three terms be   .

Product of 3 terms is 1.

Put value of a in equation 1,

The three terms of AP are .

G.P.=    , …............

Sum =120

These terms are GP with a=3 and r=3.

Hence, we have value of n as 4 to get sum of 120.

Let GP be

Given :  The sum of first three terms of a G.P. is 16

Given :  the sum of the next three terms is128.

Dividing equation (2) by (1), we have

Putting value of r =2 in equation 1,we have

Given a G.P. with a = 729 and term 64.

Given : sum of the first two terms is – 4 and the fifth term is 4 times the third term

Let first term be a and common ratio be r

If r=2, then

If r= - 2, then

Thus, required GP is       or

Let x,y, z are in G.P.

Let first term=a and common ratio = r

Dividing equation 2 by 1, we have

Dividing equation 3 by 2, we have

Equating values of  ,  we have

Thus, x,y,z are in GP

8, 88, 888, 8888… is not a GP.

It can be changed in GP by writing terms as

to n terms

Here,     is a GP.

first term =a=4

common ratio =r

To prove :      is a GP.

Thus, the above sequence is a GP with common ratio of rR.

Let first term be a and common ratio be r.

Given : the third term is greater than the first term by 9, and  the second term is greater than the by 18.

Dividing equation 2 by 1 , we get

Putting value of r , we get

Thus, four terms of GP are

To prove :

Let A  be the first term and R be common ratio.

According to the given information, we have

L.H.S :

=RHS

Thus, LHS = RHS.

Hence proved.

Given : First term =a  and n th term = b.

Common ratio = r.

To prove :

Then ,

P = product of n terms

Here,    is a AP.

Put in equation (2),

Hence proved .

Let first term =a  and common ratio = r.

Since there are n terms from (n+1) to 2n  term.

Sum of terms from (n+1) to 2n.

Thus, the required ratio  =

Thus,  the common ratio of the sum of first n terms of a G.P. to the sum of terms from   term is  .

Question:25 If a, b, c and d are in G.P. show that

If a, b, c and d are in G.P.

To prove :

RHS :

Using equation (1) and (2),

= LHS

Hence proved

Let A, B be two numbers between 3 and 81 such that series    3, A, B,81  forms a GP.

Let a=first term and common ratio =r.

For ,

The, required numbers are 9,27.

M of a and b is

Given :

Squaring both sides ,

Let there be two numbers a and b

geometric mean

According to the given condition,

.............................................................(1)

Also,

.......................................................(2)

From (1) and (2), we get

Putting the value of 'a' in (1),

Thus, the ratio is

If A and G be A.M. and G.M., respectively between two positive numbers,
Two numbers be a and b.

...................................................................1

...........................................................................2

We know

Put values from equation 1 and 2,

..................................................................3

From 1 and 3 , we have

Put value of a in equation 1, we get

Thus, numbers are

The number of bacteria in a certain culture doubles every hour.It forms GP.

Given :      a=30   and  r=2.

Thus, bacteria present at the end of the 2nd hour, 4th hour and nth hour are 120,480 and  respectively.

Given: Bank pays an annual interest rate of 10%  compounded annually.

Rs 500 amounts are deposited in the bank.

At the end of the first year, the amount

At the end of the second year, the amount

At the end of the third year, the amount

At the end of 10 years, the amount

Thus, at the end of 10 years, amount

Let roots of the quadratic equation be a and b.

According to given condition,

We know that

NCERT solutions for class 11 maths chapter 9 sequences and series-Exercise: 9.4

the series =

n th term  =

the series =

n th term  =

Thus, sum is

the series

nth term  =

Thus, the sum is

Series =

.................................

Hence, the sum is

series =

n th term  =

16th term is

Hence, the sum of the series   is 2840.

series =

=(n th term of 3,6,9,...........)(nth terms of 8,11,14,..........)

n th term  =

Hence, sum is

series =

n th term  =

nth terms is given by

nth terms are given by

This term is a GP with first term =a =2 and common ratio =r =2.

Thus, the sum is

nth terms is given by  .

## NCERT solutions for class 11 maths chapter 9 sequences and series-Miscellaneous Exercise

Let a be first term and d be common difference of AP.

Kth term of a AP is given by,

Hence, the sum of and terms of an A.P. is equal to twice the term.