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.
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.
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:2 Write the first five terms of each of the sequences in Exercises 1 to 6 whose nthterms are:
Answer:
Given :
Therefore, the required number of terms
Question:3 Write the first five terms of each of the sequences in Exercises 1 to 6 whose nthterms are:
Answer:
Given :
Therefore, required number of terms
Question:4 Write the first five terms of each of the sequences in Exercises 1 to 6 whose nth terms are:
Answer:
Given :
Therefore, the required number of terms
Question:5 Write the first five terms of each of the sequences in Exercises 1 to 6 whose nth terms are:
Answer:
Given :
Therefore, the required number of terms
Question:6 Write the first five terms of each of the sequences in Exercises 1 to 6 whose nth terms are:
Answer:
Given :
Therefore, the required number of terms
Question:7 Find the indicated terms in each of the sequences in Exercises 7 to 10 whose nth terms are:
Answer:
Put
Put n=24,
Hence, we have
Question:8 Find the indicated terms in each of the sequences in Exercises 7 to 10 whose nth terms are:
Answer:
Given :
Put n=7,
Heence, we have
Question:9 Find the indicated terms in each of the sequences in Exercises 7 to 10 whose nth terms are:
Answer:
Given :
Put n =9,
The value of
Question:10 Find the indicated terms in each of the sequences in Exercises 7 to 10 whose nth terms are:
Answer:
Given :
Put n=20,
Hence, value of
Answer:
Given :
Hence, five terms of series are
Series
Answer:
Given :
Hence, five terms of series are
Series
Answer:
Given :
Hence, five terms of series are
Series
Question:14 The Fibonacci sequence is defined by
Answer:
Given : The Fibonacci sequence is defined by
Solutions of NCERT for class 11 maths chapter 9 sequences and series-Exercise: 9.2
Question:1 Find the sum of odd integers from 1 to 2001.
Answer:
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.
Question:2 Find the sum of all natural numbers lying between 100 and 1000, which are multiples of 5.
Answer:
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.
Answer:
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.
Question:4 How many terms of the A.P. are needed to give the sum –25?
Answer:
Given : A.P. =
Given : sum = -25
Answer:
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 .
Question:6 If the sum of a certain number of terms of the A.P. 25, 22, 19, … is 116. Find the last term.
Answer:
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.
Question:7 Find the sum to n terms of the A.P., whose term is 5k + 1.
Answer:
Given :
Comparing LHS and RHS , we have
and
Putting value of d,
Question:8 If the sum of n terms of an A.P. is , where p and q are constants, find the common difference
Answer:
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.
Answer:
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
Answer:
Let first term of AP = a and common difference = d.
Then,
Given :
Now,
Thus, sum of p+q terms of AP is 0.
Question:11 Sum of the first p, q and r terms of an A.P. are a, b and c, respectively. Prove that
Answer:
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.
Question:12 The ratio of the sums of m and n terms of an A.P. is . Show that the ratio of mth and nth term is .
Answer:
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.
Question:13 If the sum of n terms of an A.P. is and its term is 164, find the value of m.
Answer:
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.
Question:14 Insert five numbers between 8 and 26 such that the resulting sequence is an A.P.
Answer:
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.
Question:15 If is the A.M. between a and b, then find the value of n.
Answer:
Given : is the A.M. between a and b.
Thus, value of n is 1.
Answer:
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.
Answer:
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.
Answer:
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.
Answer:
G.P :
first term = a
common ratio =r
the nth term of G.P
Question:2 Find the term of a G.P. whose term is 192 and the common ratio is 2.
Answer:
First term = a
common ratio =r=2
term is 192
is the term of a G.P.
Question:3 The terms of a G.P. are p, q and s, respectively. Show that
Answer:
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
Question:4 The term of a G.P. is square of its second term, and the first term is -3. Determine its term.
Answer:
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:
Answer:
Given :
n th term is given as 128.
The, 13 th term is 128.
Question:5(b) Which term of the following sequences:
Answer:
Given :
n th term is given as 729.
The, 12 th term is 729.
Question:5(c) Which term of the following sequences:
Answer:
Given :
n th term is given as
Thus, n=9.
Question:6 For what values of x, the numbers are in G.P.?
Answer:
Common ratio=r.
Thus, for ,given numbers will be in GP.
Answer:
geometric progressions is 0.15, 0.015, 0.0015, ... .....
a=0.15 , r = 0.1 , n=20
Question:9 Find the sum to indicated number of terms in each of the geometric progressions in
Answer:
The sum to the indicated number of terms in each of the geometric progressions is:
Question:12 The sum of first three terms of a G.P. is and their product is 1. Find the common ratio and the terms.
Answer:
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 .
Question:13 How many terms of G.P. , … are needed to give the sum 120?
Answer:
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.
Answer:
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
Question:15 Given a G.P. with a = 729 and term 64, determine
Answer:
Given a G.P. with a = 729 and term 64.
(Answer)
Question:16 Find a G.P. for which sum of the first two terms is – 4 and the fifth term is 4 times the third term
Answer:
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
Question:17 If the terms of a G.P. are x, y and z, respectively. Prove that x,y, z are in G.P.
Answer:
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
Question:18 Find the sum to n terms of the sequence, 8, 88, 888, 8888… .
Answer:
8, 88, 888, 8888… is not a GP.
It can be changed in GP by writing terms as
to n terms
Answer:
Here, is a GP.
first term =a=4
common ratio =r
Question:20 Show that the products of the corresponding terms of the sequences form a G.P, and find the common ratio.
Answer:
To prove : is a GP.
Thus, the above sequence is a GP with common ratio of rR.
Answer:
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
Question:22 If the terms of a G.P. are a, b and c, respectively. Prove that
Answer:
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.
Answer:
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 .
Question:24 Show that the ratio of the sum of first n terms of a G.P. to the sum of terms from term is
Answer:
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
Answer:
If a, b, c and d are in G.P.
To prove :
RHS :
Using equation (1) and (2),
= LHS
Hence proved
Question:26 Insert two numbers between 3 and 81 so that the resulting sequence is G.P.
Answer:
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.
Question:27 Find the value of n so that may be the geometric mean between a and b.
Answer:
M of a and b is
Given :
Squaring both sides ,
Question:28 The sum of two numbers is 6 times their geometric mean, show that numbers are in the ratio
Answer:
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
Question:29 If A and G be A.M. and G.M., respectively between two positive numbers, prove that the numbers are
Answer:
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
Answer:
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.
Answer:
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
Answer:
Let roots of the quadratic equation be a and b.
According to given condition,
We know that
Thus, the quadratic equation =
NCERT solutions for class 11 maths chapter 9 sequences and series-Exercise: 9.4
Question:1 Find the sum to n terms of each of the series in
Answer:
the series =
n th term =
Question:2 Find the sum to n terms of each of the series in
Answer:
the series =
n th term =
Thus, sum is
Question:3 Find the sum to n terms of each of the series
Answer:
the series
nth term =
Thus, the sum is
Question:4 Find the sum to n terms of each of the series in
Answer:
Series =
.................................
Hence, the sum is
Question:5 Find the sum to n terms of each of the series in
Answer:
series =
n th term =
16th term is
Hence, the sum of the series is 2840.
Question:6 Find the sum to n terms of each of the series
Answer:
series =
=(n th term of 3,6,9,...........)(nth terms of 8,11,14,..........)
n th term =
Hence, sum is
Question:8 Find the sum to n terms of the series in Exercises 8 to 10 whose nth terms is given by
Answer:
nth terms is given by
Question:9 Find the sum to n terms of the series in Exercises 8 to 10 whose nth terms is given by
Answer:
nth terms are given by
This term is a GP with first term =a =2 and common ratio =r =2.
Thus, the sum is
Question:10 Find the sum to n terms of the series in Exercises 8 to 10 whose nth terms is given by
Answer:
nth terms is given by .
Question:1 Show that the sum of and terms of an A.P. is equal to twice the term.
Answer:
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.