Class 11 Maths Chapter 9 Sequences and Series

Exercise 9.1

Question 1.
an = n(n + 2)
Solution:
We haven an = n(n + 2)
subtituting n = 1, 2, 3, 4, 5, we get
a1 = 9(1 + 2) = 1 x 3 = 3
a2 = 2(2 + ) = 2 x 4 = 8
a3 = 3(3 + 2) = 3 x 5 = 15
a4 = 4(4 + 2) = 4 x 6 = 24
a5 = 5(5 + 2) = 5 x 7 = 35
∴ The first five terms are 3, 8, 15, 24, 35.

Question 2.
an = 
Solution:

Question 3.
an = 2n
Solution:

Question 4.
an = 
Solution:

Question 5.
an = (- 1)n-1 5n+1
Solution:
We have, an = (- 1)n-1 5n+1
Substituting n = 1, 2, 3, 4, 5, we get
a1 =(-1)1-1 51+1 = (-1)° 52 = 25
a2 =(-1)2-1 52+1 = (-1)1 53 = 125
a3 =(-1)3-1 53+1 = (-1)2 54 = 625
a4 =(-1)4-1 54+1 = (-1)3 55 = -3125
a5 =(-1)5-1 55+1 = (-1)4 56 = 15625
∴ The first five terms are 25, – 125, 625, -3125, 15625.

Question 6.
an = n
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:

Question 7.
an = 4n – 3; a17, a24
Solution:
We have an = 4n – 3

Question 8.
an = ; a7
Solution:
We have, an = ; a7

Question 9.
an = (-1)n – 1 n3; a9
Solution:
We have, an = (-1)n – 1 n3

Question 10.
an = ; a 20
Solution:
We have, an = 

Write the first five terms of each of the sequences in Exercises 11 to 13 and obtain the corresponding series:

Question 11.
a1 = 3, an = 3an-1+2 for all n>1
Solution:
We have given a1 = 3, an = 3an-1+2
⇒ a1 = 3, a2 = 3a1 + 2 = 3.3 + 2 = 9 + 2 = 11,
a3 = 3a2 + 2 = 3.11 + 2 = 33 + 2 = 35,
a4 = 3a3 + 2 = 3.35 + 2 = 105 + 2 = 107,
a5 = 3a4 + 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 + ………..

Question 12.
a1 = -1, an = , 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

 Question 13.
a1 = a2 = 2, an = an-1 – 1, n >2
Solution:
We have given a1 = a2 = 2, an = an-1 – 1, n >2
a1 = 2, a2 = 2, a3= a2 – 1 = 2 – 1 = 1,
a4 = a3 – 1 = 1 – 1 = 0 and a5 = a4 – 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) + ……

Question 14.
Find Fibonacci sequence is defined by 1 = a1 = a2 and an = an-1 + an-2, n > 2
Find , for n = 1, 2, 3, 4, 5
Solution:
We have,

Exercise 9.2

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 = 3-1 = 2 and last term l = 2001
∴ l = a + (n-1 )d ⇒ 2001 = 1 + (n -1)2 ⇒ 2001 = 1 + 2n – 2 ⇒ 2n = 2001 + 1

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

Question 3.
In an A.P., the first term is 2 and the sum of the first five terms is one-fourth of the next five terms. Show that 20th term is -112.
Solution:
Let a = 2 be the first term and d be the common difference.

Question 4.
How many terms of the A.P. -6, , -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

Question 5.
In an A.P., if pth term is  and qth term is  prove that the sum of first pq terms is , where p±q.
Solution:
Let a be the first term & d be the common difference of the A.P., then

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, Sn = 116

Question 7.
Find the sum to n terms of the A.P., whose kth term is 5k + 1.
Solution:
We have ak = 5k +1
By substituting the value of k = 1, 2, 3 and 4,
we get

Question 8.
If the sum of n terms of an A.P. is (pn + qn2), where p and q are constants, find the common difference.
Solution:
We have Sn = pn + qn2, where S„ be the sum of n terms.

Question 9.
The sums of n terms of two arithmetic progressions are in the ratio 5n + 4 : 9n + 6. Find the ratio of their 18th terms.
Solution:
Let a1, a2 & d1 d2 be the first terms & common differences of the two arithmetic progressions respectively. According to the given condition, we have

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

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

Question 12.
The ratio of the sums of m and n terms of an A.P. is m2: n2. Show that the ratio of mth and nth term is (2m -1): (2n -1).
Solution:
Let the first term be a & common difference be d. Then

Question 13.
If the sum of n terms of an A.P. is 3n2 + 5n and its mth term is 164, find the value of m.
Solution:
We have Sn = 3n2 + 5n, where Sn be the sum of n terms.

Question 14.
Insert five numbers between 8 and 26 such that the resulting sequence is an A.P.
Solution:
Let A1, A2, A3, A4, A5 be numbers between 8 and 26 such that 8, A1, A2, A3, A4, A5, 26 are in A.P., Here a = 8, l = 26, n = 7

Question 15.
If  is the A.M. between a and b, then find the value of n.
Solution:
We have 

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 7thand (m – 1 )th numbers is 5:9. Find the value of m.
Solution:
Let the sequence be 1, A1, A2, ……… Am, 31 Then 31 is (m + 2)th term, a = 1, let d be the common difference

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
∴ an= a + 29d = 100 + 29(5) = 100 + 145 = 245
Hence he will pay Rs. 245 in 30th instalment.

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.

Exercise 9.3

Question 1.
Find the 20th and nth terms of the G.P. , ….
Solution:

Question 2.
Find the 12th term of a G.P. whose 8th term is 192 and the common ratio is 2.
Solution:
We have, as = 192, r = 2

Question 3.
The 5th, 8th and 11th terms of a G.P. are p, q and s, respectively. Show that q2 = ps.
Solution:
We are given

Question 4.
The 4th term of a G.P. is square of its second term, and the first term is – 3. Determine its 7th term.
Solution:
We have a= -3, a4 = (a2)2

Question 5.
Which term of the following sequences:


Solution:

Question 6.
For what values of x, the numbers , x,  are in G.P.?
Solution:

Find the sum to indicated number of terms in each of the geometric progressions in Exercises 7 to 10:

Question 7.
0.14, 0.015, 0.0015, …. 20 items.
Solution:
In the given G.P.

Question 8.
, …. n terms
Solution:
In the given G.P.

Question 9.
1, -a, a2,- a3 … n terms (if a ≠ -1)
Solution:
In the given G.P.. a = 1, r = -a

Question 10.
x3, x5, 7, ….. n terms (if ≠±1).
Solution:
In the given G.P., a = x3, r = x2

Question 11.
Evaluate 
Solution:

Question 12.
The sum of first three terms of a G.P. is  and 10 their product is 1. Find the common ratio and the terms.
Solution:
Let the first three terms of G.P. be ,a,ar, where a is the first term and r is the common ratio.

Question 13.
How many terms of G.P. 3, 32, 33, … are needed to give the sum 120?
Solution:
Let n be the number of terms we needed. Here a = 3, r = 3, Sn = 120

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 a1, a2, a3, a4, a5, a6 be the first six terms of the G.P.

Question 15.
Given a G.P. with a = 729 and 7th term 64, determine S7.
Solution:
Let a be the first term and the common ratio be r.

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 a1 a2 be first two terms and a3 a5 be third and fifth terms respectively.
According to question

Question 17.
If the 4th, 10th and 16th terms of a G.P. are x, y and z respectively. Prove that x, y, z are in G.P.
Solution:
Let a be the first term and r be the common ratio, then according to question

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 Sn= 8 +88 + 888 + 8888 + to n terms

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, 
Solution:
On multiplying the corresponding terms of sequences, we get 256, 128, 64, 32 and 16, which forms a G.P. of 5 terms

Question 20.
Show that the products of the corresponding terms of the sequences a, ar, ar2, ………… arn-1 and A, AR, AR2, …….. , ARn-1 form a G.P., and find the common ratio.
Solution:
On multiplying the corresponding terms, we get aA, aArR, aAr2R2,…… aArn-1Rn-1. We can see that this new sequence is G.P. with first term aA & the common ratio rR.

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 4th by 18.
Solution:
Let the four numbers forming a G.P. be a, ar, ar2, ar3
According to question,

Question 22.
If the pth, qth and rth terms of a G.P. are a, b and c, respectively. Prove that aq-r br-p cp-q = 1.
Solution:
Let A be the first term and R be the common ratio, then according to question

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 P2 – (ab)n.
Solution:
Let r be the common ratio of the given G.P., then b = nth term = arn-1

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 
Solution:
Let the G.P. be a, ar, ar2, ……
Sum of first n terms = a + ar + ……. + arn-1

Question 25.
If a, b,c and d are in G.P., show that
(a2 + b2 + c2) (b2 + c2 + d2) = (ab + bc + cd)2
Solution:
We have a, b, c, d are in G.P.
Let r be a common ratio, then

Question 26.
Insert two numbers between 3 and 81 so that the resulting sequence is G.P.
Solution:
Let G1, G2 be two numbers between 3 and 81 such that 3, G1 G2,81 is a G.P.

Question 27.
Find the value of n so that  may be the geometric mean between a and b.
Solution:

Question 28.
The sum of two numbers is 6 times their geometric mean, show that numbers are in the ratio .
Solution:
Let a and b be the two numbers such that a + b = 6 

Question 29.
If A and G be A.M. and G.M., respectively between two positive numbers, prove that the numbers are .
Solution:
Let a and b be the numbers such that A, G are A.M. and G.M. respectively between them.

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, …….

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

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.

Exercise 9.4

Find the sunt to n terms of each of the series in Exercises 1 to 7.

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

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

Question 3.
3 x 12 + 5 x 22 + 7 x 32 + …..
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) 12, 22, 32, ………………….
Now the nth term of sum is an = (nth term of the sequence formed by first A.P.) x (nth term of the sequence formed by second A.P.) = (2 n + 1) x n2 = 2n3 + n2 Hence, the sum to n terms is,

Question 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

Question 5.
52 + 62 + 72 + ………….. + 202
Solution:
The given series can be written in the following way

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 = (nth term of the sequence formed by first A.P.) x (nth term of the sequence formed by second A.P.)

Question 7.
12 + (12 + 22) + (12 + 22 + 32) + ………….
Solution:
In the given series
an = 12 + 22 + …………….. + n2

Find the sum to n terms of the series in Exercises 8 to 10 whose nth terms is given by

Question 8.
n(n + 1)(n + 4)
Solution:
We have

Question 9.
n2 + 2n
Solution:
We have an = n2 + 2n
Hence, the sum to n terms is,

Question 10.
(2n – 1)2
Solution:
We have