NCERT Solutions  Class 11 Maths Chapter 9  Sequences and series
CBSE Solutions  Maths Class 11
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NCERT  Class 11 Maths
Book:  National Council of Educational Research and Training (NCERT) 

Board:  Central Board of Secondary Education (CBSE) 
Class:  11th 
Subject:  Maths 
Chapter:  9 
Chapters Name:  Sequences and series 
Medium:  English 
Sequences and series  Class 11 Maths  NCERT Books Solutions
NCERT Solutions for Class 11 Maths Chapter 9 Sequences and Series Ex 9.1
Ex 9.1 Class 11 Maths Question 1.
a_{n} = n(n + 2)
Solution.
We haven a_{n} = n(n + 2)
subtituting n = 1, 2, 3, 4, 5, we get
a_{1} = 9(1 + 2) = 1 x 3 = 3
a_{2} = 2(2 + ) = 2 x 4 = 8
a_{3} = 3(3 + 2) = 3 x 5 = 15
a_{4} = 4(4 + 2) = 4 x 6 = 24
a_{5} = 5(5 + 2) = 5 x 7 = 35
∴ The first five terms are 3, 8, 15, 24, 35.
Ex 9.1 Class 11 Maths Question 2.
a_{n} = \(\frac { n }{ n+1 } \)
Solution.
Ex 9.1 Class 11 Maths Question 3.
a_{n} = 2^{n}
Solution.
Ex 9.1 Class 11 Maths Question 4.
a_{n} = \(\frac { 2n3 }{ 6 } \)
Solution.
Ex 9.1 Class 11 Maths Question 5.
a_{n} = ( 1)^{n1} 5^{n+1}
Solution.
We have, a_{n} = ( 1)^{n1} 5^{n+1}
Substituting n = 1, 2, 3, 4, 5, we get
a_{1} =(1)^{11} 5^{1+1} = (1)^{°} 5^{2} = 25
a_{2} =(1)^{21} 5^{2+1} = (1)^{1} 5^{3} = 125
a_{3} =(1)^{31} 5^{3+1} = (1)^{2} 5^{4} = 625
a_{4} =(1)^{41} 5^{4+1} = (1)^{3} 5^{5} = 3125
a_{5} =(1)^{51} 5^{5+1} = (1)^{4} 5^{6} = 15625
∴ The first five terms are 25, – 125, 625, 3125, 15625.
Ex 9.1 Class 11 Maths Question 6.
a_{n} = n\(\frac { { n }^{ 2 }+5 }{ 4 } \)
Solution.
Substituting n = 1, 2, 3, 4, 5, we get
Find the indicated terms in each of the sequences in Exercises 7 to 10 whose nth terms are:
Ex 9.1 Class 11 Maths Question 7.
a_{n} = 4n – 3; a_{17}, a_{24}
Solution.
We have a_{n} = 4n – 3
Ex 9.1 Class 11 Maths Question 8.
a_{n} = \(\frac { { n }^{ 2 } }{ { 2 }^{ n } } \); a_{7}
Solution.
We have, a_{n} = \(\frac { { n }^{ 2 } }{ { 2 }^{ n } } \); a_{7}
Ex 9.1 Class 11 Maths Question 9.
a_{n} = (1)^{n – 1} n^{3}; a_{9}
Solution.
We have, a_{n} = (1)^{n – 1} n^{3}
Ex 9.1 Class 11 Maths Question 10.
a_{n} = \(\frac { n(n2) }{ n+3 } \); a _{20}
Solution.
We have, a_{n} = \(\frac { n(n2) }{ n+3 } \)
Write the first five terms of each of the sequences in Exercises 11 to 13 and obtain the corresponding series:
Ex 9.1 Class 11 Maths Question 11.
a_{1} = 3, a_{n} = 3a_{n1}+2 for all n>1
Solution.
We have given a_{1} = 3, a_{n} = 3a_{n1}+2
⇒ a_{1} = 3, a_{2} = 3a_{1} + 2 = 3.3 + 2 = 9 + 2 = 11,
a_{3} = 3a_{2} + 2 = 3.11 + 2 = 33 + 2 = 35,
a_{4} = 3a_{3} + 2 = 3.35 + 2 = 105 + 2 = 107,
a_{5} = 3a_{4} + 2 = 3.107 + 2 = 321 + 2 = 323,
Hence, the first five terms of the sequence are 3, 11, 35, 107, 323.
The corresponding series is 3 + 11 + 35 + 107 + 323 + ………..
Ex 9.1 Class 11 Maths Question 12.
a_{1} = 1, a_{n} = \(\frac { { a }_{ n }1 }{ n } \), n ≥ 2
Solution.
We have given
Hence the first five terms of the given sequence are – 1, 1/2, 1/6, 1/24, 1/120.
The corresponding series is
Ex 9.1 Class 11 Maths Question 13.
a_{1} = a_{2} = 2, a_{n} = a_{n1} – 1, n >2
Solution.
We have given a_{1} = a_{2} = 2, a_{n} = a_{n1} – 1, n >2
a_{1} = 2, a_{2} = 2, a_{3}= a_{2} – 1 = 2 – 1 = 1,
a_{4} = a_{3} – 1 = 1 – 1 = 0 and a_{5} = a_{4} – 1 = 0 – 1 = 1
Hence the first five terms of the sequence are 2, 2, 1, 0, 1
The corresponding series is
2 + 2 + 1 + 0 + (1) + ……
Ex 9.1 Class 11 Maths Question 14.
Find Fibonacci sequence is defined by 1 = a_{1} = a_{2} and a_{n} = a_{n1} + a_{n2}, n > 2
Find \(\frac { { a }_{ n }+1 }{ { a }_{ n } } \), for n = 1, 2, 3, 4, 5
Solution.
We have,
NCERT Solutions for Class 11 Maths Chapter 9 Sequences and Series Ex 9.2
Ex 9.2 Class 11 Maths Question 1.
Find the sum of odd integers from 1 to 2001.
Solution.
We have to find 1 + 3 + 5 + ……….. + 2001
This is an A.P. with first term a = 1, common difference d = 31 = 2 and last term l = 2001
∴ l = a + (n1 )d ⇒ 2001 = 1 + (n 1)2 ⇒ 2001 = 1 + 2n – 2 ⇒ 2n = 2001 + 1
Ex 9.2 Class 11 Maths Question 2.
Find the sum of all natural numbers lying between 100 and 1000, which are multiples of 5.
Solution.
We have to find 105 +110 +115 + ……..+ 995
This is an A.P. with first term a = 105, common difference d = 110 105 = 5 and last term 1 = 995
Ex 9.2 Class 11 Maths Question 3.
In an A.P., the first term is 2 and the sum of the first five terms is onefourth of the next five terms. Show that 20th term is 112.
Solution.
Let a = 2 be the first term and d be the common difference.
Ex 9.2 Class 11 Maths Question 4.
How many terms of the A.P. 6, \(\frac { 11 }{ 2 } \), 5, ……. are needed to give the sum – 25 ?
Solution.
Let a be the first term and d be the common difference of the given A.P., we have
Ex 9.2 Class 11 Maths Question 5.
In an A.P., if pth term is \(\frac { 1 }{ q } \) and qth term is \(\frac { 1 }{ p } \) prove that the sum of first pq terms is \(\frac { 1 }{ 2 } \left( pq+1 \right) \), where p±q.
Solution.
Let a be the first term & d be the common difference of the A.P., then
Ex 9.2 Class 11 Maths Question 6.
If the sum of a certain number of terms of the A.P. 25, 22, 19, …. is 116. Find the last term.
Solution.
Let a be the first term and d be the common difference.
We have a = 25, d = 22 – 25 = 3, S_{n} = 116
Ex 9.2 Class 11 Maths Question 7.
Find the sum to n terms of the A.P., whose k^{th} term is 5k + 1.
Solution.
We have a_{k} = 5k +1
By substituting the value of k = 1, 2, 3 and 4,
we get
Ex 9.2 Class 11 Maths Question 8.
If the sum of n terms of an A.P. is (pn + qn^{2}), where p and q are constants, find the common difference.
Solution.
We have S_{n} = pn + qn^{2}, where S„ be the sum of n terms.
Ex 9.2 Class 11 Maths Question 9.
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.
Solution.
Let a_{1}, a_{2} & d_{1} d_{2} be the first terms & common differences of the two arithmetic progressions respectively. According to the given condition, we have
Ex 9.2 Class 11 Maths Question 10.
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.
Solution.
Let the first term be a and common difference be d.
According to question
Ex 9.2 Class 11 Maths Question 11.
Sum of the first p, q and r terms of an A.P. are a, b and c, respectively.
Prove that
Solution.
Let the first term be A & common difference be D. We have
Ex 9.2 Class 11 Maths Question 12.
The ratio of the sums of m and n terms of an A.P. is m^{2}: n^{2}. Show that the ratio of m^{th} and n^{th} term is (2m 1): (2n 1).
Solution.
Let the first term be a & common difference be d. Then
Ex 9.2 Class 11 Maths Question 13.
If the sum of n terms of an A.P. is 3n^{2} + 5n and its mth term is 164, find the value of m.
Solution.
We have S_{n} = 3n^{2} + 5n, where Sn be the sum of n terms.
Ex 9.2 Class 11 Maths Question 14.
Insert five numbers between 8 and 26 such that the resulting sequence is an A.P.
Solution.
Let A_{1}, A_{2}, A_{3}, A_{4}, A_{5} be numbers between 8 and 26 such that 8, A_{1}, A_{2}, A_{3}, A_{4}, A_{5}, 26 are in A.P., Here a = 8, l = 26, n = 7
Ex 9.2 Class 11 Maths Question 15.
If \(\frac { { a }^{ n }+{ b }^{ n } }{ { a }^{ n1 }+{ b }^{ n1 } } \) is the A.M. between a and b, then find the value of n.
Solution.
We have \(\frac { { a }^{ n }+{ b }^{ n } }{ { a }^{ n1 }+{ b }^{ n1 } } =\frac { a+b }{ 2 } \)
Ex 9.2 Class 11 Maths Question 16.
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.
Solution.
Let the sequence be 1, A_{1}, A_{2}, ……… A_{m}, 31 Then 31 is (m + 2)^{th} term, a = 1, let d be the common difference
Ex 9.2 Class 11 Maths Question 17.
A man starts repaying a loan as first instalment of Rs. 100. If he increases the instalment by Rs. 5 every month, what amount he will pay in the 30th instalment?
Solution.
Here, we have an A.P. with a = 100 and d = 5
∴ a_{n}= a + 29d = 100 + 29(5) = 100 + 145 = 245
Hence he will pay Rs. 245 in 30^{th} instalment.
Ex 9.2 Class 11 Maths Question 18.
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.
Solution.
Let there are n sides of a polygon.
NCERT Solutions for Class 11 Maths Chapter 9 Sequences and Series Ex 9.3
Ex 9.3 Class 11 Maths Question 1.
Find the 20^{th} and n^{th} terms of the G.P. \(\frac { 5 }{ 2 } ,\frac { 5 }{ 4 } ,\frac { 5 }{ 8 } \), ….
Solution.
Ex 9.3 Class 11 Maths Question 2.
Find the 12^{th} term of a G.P. whose 8^{th} term is 192 and the common ratio is 2.
Solution.
We have, a_{s} = 192, r = 2
Ex 9.3 Class 11 Maths Question 3.
The 5^{th}, 8^{th} and 11th terms of a G.P. are p, q and s, respectively. Show that q^{2} = ps.
Solution.
We are given
Ex 9.3 Class 11 Maths Question 4.
The 4th term of a G.P. is square of its second term, and the first term is – 3. Determine its 7^{th} term.
Solution.
We have a= 3, a_{4} = (a_{2})^{2}
Ex 9.3 Class 11 Maths Question 5.
Which term of the following sequences:
Solution.
Ex 9.3 Class 11 Maths Question 6.
For what values of x, the numbers \(\frac { 2 }{ 7 } \), x, \(\frac { 7 }{ 2 } \) are in G.P.?
Solution.
Find the sum to indicated number of terms in each of the geometric progressions in Exercises 7 to 10:
Ex 9.3 Class 11 Maths Question 7.
0.14, 0.015, 0.0015, …. 20 items.
Solution.
In the given G.P.
Ex 9.3 Class 11 Maths Question 8.
\(\sqrt { 7 } ,\quad \sqrt { 21 } ,\quad 3\sqrt { 7 } \), …. n terms
Solution.
In the given G.P.
Ex 9.3 Class 11 Maths Question 9.
1, a, a^{2}, a^{3} … n terms (if a ≠ 1)
Solution.
In the given G.P.. a = 1, r = a
Ex 9.3 Class 11 Maths Question 10.
x^{3}, x^{5}, ^{7}, ….. n terms (if ≠±1).
Solution.
In the given G.P., a = x^{3}, r = x^{2}
Ex 9.3 Class 11 Maths Question 11.
Evaluate
Solution.
Ex 9.3 Class 11 Maths Question 12.
The sum of first three terms of a G.P. is \(\frac { 39 }{ 10 } \) and 10 their product is 1. Find the common ratio and the terms.
Solution.
Let the first three terms of G.P. be \(\frac { a }{ r } \),a,ar, where a is the first term and r is the common ratio.
Ex 9.3 Class 11 Maths Question 13.
How many terms of G.P. 3, 3^{2}, 3^{3}, … are needed to give the sum 120?
Solution.
Let n be the number of terms we needed. Here a = 3, r = 3, S_{n} = 120
Ex 9.3 Class 11 Maths Question 14.
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.
Solution.
Let a_{1}, a_{2}, a_{3}, a_{4}, a_{5}, a_{6} be the first six terms of the G.P.
Ex 9.3 Class 11 Maths Question 15.
Given a G.P. with a = 729 and 7^{th} term 64, determine S_{7}.
Solution.
Let a be the first term and the common ratio be r.
Ex 9.3 Class 11 Maths 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.
Solution.
Let a_{1} a_{2} be first two terms and a_{3} a_{5} be third and fifth terms respectively.
According to question
Ex 9.3 Class 11 Maths Question 17.
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.
Solution.
Let a be the first term and r be the common ratio, then according to question
Ex 9.3 Class 11 Maths Question 18.
Find the sum to n terms of the sequence, 8, 88, 888, 8888 ………
Solution.
This is not a G.P., however we can relate it to a G.P. by writing the terms as S_{n}= 8 +88 + 888 + 8888 + to n terms
Ex 9.3 Class 11 Maths Question 19.
Find the sum of the products of the corresponding terms of the sequences 2, 4, 8, 16, 32 and 128, 32, 8, 2, \(\frac { 1 }{ 2 } \)
Solution.
On multiplying the corresponding terms of sequences, we get 256, 128, 64, 32 and 16, which forms a G.P. of 5 terms
Ex 9.3 Class 11 Maths Question 20.
Show that the products of the corresponding terms of the sequences a, ar, ar^{2}, ………… ar^{n1} and A, AR, AR^{2}, …….. , AR^{n1} form a G.P., and find the common ratio.
Solution.
On multiplying the corresponding terms, we get aA, aArR, aAr^{2}R^{2},…… aAr^{n1}R^{n1}. We can see that this new sequence is G.P. with first term aA & the common ratio rR.
Ex 9.3 Class 11 Maths Question 21.
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.
Solution.
Let the four numbers forming a G.P. be a, ar, ar^{2}, ar^{3}
According to question,
Ex 9.3 Class 11 Maths Question 22.
If the p^{th}, q^{th} and r^{th} terms of a G.P. are a, b and c, respectively. Prove that a^{qr} b^{rp} c^{pq} = 1.
Solution.
Let A be the first term and R be the common ratio, then according to question
Ex 9.3 Class 11 Maths Question 23.
If the first and the nth term of a G.P. are a and b, respectively, and if P is the product of n terms, prove that P^{2} – (ab)^{n}.
Solution.
Let r be the common ratio of the given G.P., then b = n^{th} term = ar^{n1}
Ex 9.3 Class 11 Maths Question 24.
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 \(\frac { 1 }{ { r }^{ n } } \)
Solution.
Let the G.P. be a, ar, ar^{2}, ……
Sum of first n terms = a + ar + ……. + ar^{n1}
Ex 9.3 Class 11 Maths Question 25.
If a, b,c and d are in G.P., show that
(a^{2} + b^{2} + c^{2}) (b^{2} + c^{2} + d^{2}) = (ab + bc + cd)^{2}
Solution.
We have a, b, c, d are in G.P.
Let r be a common ratio, then
Ex 9.3 Class 11 Maths Question 26.
Insert two numbers between 3 and 81 so that the resulting sequence is G.P.
Solution.
Let G_{1}, G_{2} be two numbers between 3 and 81 such that 3, G_{1} G_{2},81 is a G.P.
Ex 9.3 Class 11 Maths Question 27.
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.
Solution.
Ex 9.3 Class 11 Maths Question 28.
The sum of two numbers is 6 times their geometric mean, show that numbers are in the ratio \(\left( 3+2\sqrt { 2 } \right) :\left( 32\sqrt { 2 } \right) \).
Solution.
Let a and b be the two numbers such that a + b = 6 \(\sqrt { ab } \)
Ex 9.3 Class 11 Maths Question 29.
If A and G be A.M. and G.M., respectively between two positive numbers, prove that the numbers are \(A\pm \sqrt { \left( A+G \right) \left( AG \right) } \).
Solution.
Let a and b be the numbers such that A, G are A.M. and G.M. respectively between them.
Ex 9.3 Class 11 Maths Question 30.
The number of bacteria in a certain culture doubles every hour. If there were 30 bacteria present in the culture originally, how many bacteria will be present at the end of 2nd hour, 4th hour and nth hour?
Solution.
There were 30 bacteria present in the culture originally and it doubles every hour. So, the number of bacteria at the end of successive hours form the G.P. i.e., 30, 60, 120, 240, …….
Ex 9.3 Class 11 Maths Question 31.
What will Rs. 500 amounts to in 10 years after its deposit in a bank which pays annual interest rate of 10% compounded annually?
Solution.
We have, Principal value = Rs. 500 Interest rate = 10% annually
Ex 9.3 Class 11 Maths Question 32.
If A.M. and G.M. of roots of a quadratic equation are 8 and 5, respectively, then obtain the quadratic equation.
Solution.
Let α & β be the roots of a quadratic equation such that A.M. & G.M. of α, β are 8 and 5 respectively.
NCERT Solutions for Class 11 Maths Chapter 9 Sequences and Series Ex 9.4
Find the sunt to n terms of each of the series in Exercises 1 to 7.
Ex 9.4 Class 11 Maths Question 1.
1 x 2 + 2 x 3 + 3 x 4 + 4 x 5 + ………
Solution.
In the given series, there is a sum of multiple of corresponding terms of two A.P’s. The two A.P’s are
Ex 9.4 Class 11 Maths Question 2.
1 x 2 x 3 + 2 x 3 x 4 + 3 x 4 x 5 + ……
Solution.
In the given series, there is a sum of multiple of corresponding terms of two A.P’s. The three A.P’s are
Ex 9.4 Class 11 Maths Question 3.
3 x 1^{2} + 5 x 2^{2} + 7 x 3^{2} + …..
Solution.
In the given series there is sum of multiple of corresponding terms of two A.P’s. The two A.P’s are
(i) 3, 5, 7, …………… and
(ii) 1^{2}, 2^{2}, 3^{2}, ………………….
Now the n^{th} term of sum is an = (nth term of the sequence formed by first A.P.) x (n^{th} term of the sequence formed by second A.P.) = (2 n + 1) x n^{2} = 2n^{3} + n^{2} Hence, the sum to n terms is,
Ex 9.4 Class 11 Maths Question 4.
\(\frac { 1 }{ 1\times 2 } +\frac { 1 }{ 2\times 3 } +\frac { 1 }{ 3\times 4 } +\) …….
Solution.
In the given series there is sum of multiple of corresponding terms of two A.P’s. The two A.P’s are
Ex 9.4 Class 11 Maths Question 5.
5^{2} + 6^{2} + 7^{2} + ………….. + 20^{2}
Solution.
The given series can be written in the following way
Ex 9.4 Class 11 Maths Question 6.
3 x 8 + 6 x 11 + 9 x 25 + ………….
Solution.
In the given series, there is sum of multiple of corresponding terms of two A.P/s. The two A.P/s are
(i) 3, 6, 9, ………….. and
(ii) 8, 11, 14, ……………….
Now the nth term of sum is an = (n^{th} term of the sequence formed by first A.P.) x (n^{th} term of the sequence formed by second A.P.)
Ex 9.4 Class 11 Maths Question 7.
1^{2} + (1^{2} + 2^{2}) + (1^{2} + 2^{2} + 3^{2}) + ………….
Solution.
In the given series
a_{n} = 1^{2} + 2^{2} + …………….. + n^{2}
Find the sum to n terms of the series in Exercises 8 to 10 whose n^{th} terms is given by
Ex 9.4 Class 11 Maths Question 8.
n(n + 1)(n + 4)
Solution.
We have
Ex 9.4 Class 11 Maths Question 9.
n^{2} + 2^{n}
Solution.
We have a_{n} = n^{2} + 2^{n}
Hence, the sum to n terms is,
Ex 9.4 Class 11 Maths Question 10.
(2n – 1)^{2}
Solution.
We have
NCERT Class 11 Maths
Class 11 Maths Chapters  Maths Class 11 Chapter 9
Chapterwise NCERT Solutions for Class 11 Maths

NCERT Solutions For Class 11 Maths Chapter 1 Sets
NCERT Solutions For Class 11 Maths Chapter 2 Functions and relations
NCERT Solutions For Class 11 Maths Chapter 3 Trigonometric functions
NCERT Solutions For Class 11 Maths Chapter 4 Principle of mathematical induction
NCERT Solutions For Class 11 Maths Chapter 5 Quadratic equations and complex numbers
NCERT Solutions For Class 11 Maths Chapter 6 Linear Inequalities
NCERT Solutions For Class 11 Maths Chapter 7 Permutations and combinations
NCERT Solutions For Class 11 Maths Chapter 8 Binomial Theorem
NCERT Solutions For Class 11 Maths Chapter 9 Sequences and series
NCERT Solutions For Class 11 Maths Chapter 10 Straight lines
NCERT Solutions For Class 11 Maths Chapter 11 Conic sections
NCERT Solutions For Class 11 Maths Chapter 12 Introduction to threedimensional geometry
NCERT Solutions For Class 11 Maths Chapter 13 Limits and derivatives
NCERT Solutions For Class 11 Maths Chapter 14 Mathematical Reasoning
NCERT Solutions For Class 11 Maths Chapter 15 Statistics
NCERT Solutions For Class 11 Maths Chapter 16 Probability
NCERT Solutions for Class 12 All Subjects  NCERT Solutions for Class 10 All Subjects 
NCERT Solutions for Class 11 All Subjects  NCERT Solutions for Class 9 All Subjects 
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