# MCQ Questions for Class 11 Maths Chapter 9 ﻿Sequences and Series with Answers

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## Sequences and Series Class 11 MCQs Questions with Answers

Solving the MCQ Questions of Sequences and Series Class 11 with answers can help you understand the concepts better.

A sequence is a list of items/objects which have been arranged in a sequential way.

series can be highly generalized as the sum of all the terms in a sequence. However, there has to be a definite relationship between all the terms of the sequence.

Question 1.
Let Tr be the rth term of an A.P. whose first term is a and the common difference is d. If for some positive integers m, n, m ≠ n, T m = 1/n and T n = 1/m then (a-d) equals to
(a) 0
(b) 1
(c) 1/mn
(d) 1/m + 1/n

Given the first term is a and the common difference is d of the AP
Now, Tm = 1/n
⇒ a + (m – 1)d = 1/n ………… 1
and Tn = 1/m
⇒ a + (n – 1)d = 1/m ………. 2
From equation 2 – 1, we get
(m – 1)d – (n – 1)d = 1/n – 1/m
⇒ (m – n)d = (m – n)/mn
⇒ d = 1/mn
From equation 1, we get
a + (m – 1)/mn = 1/n
⇒ a = 1/n – (m – 1)/mn
⇒ a = {m – (m – 1)}/mn
⇒ a = {m – m + 1)}/mn
⇒ a = 1/mn
Now, a – d = 1/mn – 1/mn
⇒ a – d = 0

Question 2.
The first term of a GP is 1. The sum of the third term and fifth term is 90. The common ratio of GP is
(a) 1
(b) 2
(c) 3
(d) 4

Let first term of the GP is a and common ratio is r.
3rd term = ar²
5th term = ar4
Now
⇒ ar² + ar4 = 90
⇒ a(r² + r4) = 90
⇒ r² + r4 = 90
⇒ r² × (r² + 1) = 90
⇒ r² (r² + 1) = 3² × (3² + +1)
⇒ r = 3
So the common ratio is 3

Question 3.
If a is the first term and r is the common ratio then the nth term of GP is
(a) (ar)n-1
(b) a × rⁿ
(c) a × rn-1
(d) None of these

Given, a is the first term and r is the common ratio.
Now, nth term of GP = a × rn-1

Question 4.
The sum of odd integers from 1 to 2001 is
(a) 10201
(b) 102001
(c) 100201
(d) 1002001

The odd numbers from 1 to 2001 are:
1, 3, 5, ………, 2001
This froms an AP
where first term a = 1
Common difference d = 3 – 1 = 2
last term l = 2001
Let number of terms = n
Now, l = a + (n – 1)d
⇒ 2001 = 1 + (n – 1)2
⇒ 2001 – 1 = (n – 1)2
⇒ 2(n – 1) = 2000
⇒ n – 1 = 2000/2
⇒ n – 1 = 1000
⇒ n = 1001
Now, sum = (n/2) × (a + 1)
= (1001/2) × (1 + 2001)
= (1001/2) × 2002
= 1001 × 1001
= 1002001
So, the sum of odd integers from 1 to 2001 is 1002001

Question 5.
If a, b, c are in AP and x, y, z are in GP then the value of xb-c × yc-a × za-b is
(a) 0
(b) 1
(c) -1
(d) None of these

Given, a, b, c are in AP
⇒ 2b = a + c ………. 1
and x, y, z are in GP
⇒ y² = xz ……….. 2
Now, xb-c × yc-a × za-b = xb-c × (√xz)c-a × za-b
= xb-c × x(c-a)/2 × z(c-a)/2 × za-b
= xb-c + x(c-a)/2 × z(c-a)/2+ a -b
= x2b+(c+a) × z(c+a)-2b
= x° × z°
= 1
So, the value of xb-c × yc-a × za-b is 1

Question 6.
An example of geometric series is
(a) 9, 20, 21, 28
(b) 1, 2, 4, 8
(c) 1, 2, 3, 4
(d) 3, 5, 7, 9

Answer: (b) 1, 2, 4, 8
1, 2, 4, 8 is the example of geometric series
Here common ratio = 2/1 = 4/2 = 8/4 = 2

Question 7.
Three numbers from an increasing GP of the middle number is doubled, then the new numbers are in AP. The common ratio of the GP is
(a) 2
(b) √3
(c) 2 + √3
(d) 2 – √3

Given that three numbers from an increasing GP
Let the 3 number are: a, ar, ar² (r > 1)
Now, according to question,
a, 2ar, ar² are in AP
So, 2ar – a = ar² – 2ar
⇒ a(2r – 1) = a(r² – 2r)
⇒ 2r – 1 = r² – 2r
⇒ r² – 2r – 2r + 1 = 0
⇒ r² – 4r + 1 = 0
⇒ r = [4 ± √{16 – 4 × 1 × 1}]/2
⇒ r = [4 ± √{16 – 4}]/2
⇒ r = {4 ± √12}/2
⇒ r = {4 ± 2√3}/2
⇒ r = {2 ± √3}
Since r > 1
So, the common ratio of the GP is (2 + √3)

Question 8.
An arithmetic sequence has its 5th term equal to 22 and its 15th term equal to 62. Then its 100th term is equal to
(a) 410
(b) 408
(c) 402
(d) 404

Let ais the first term and d is the common difference of the AP
Given,
a5 = a + (5 – 1)d = 22
⇒ a + 4d = 22 ………….1
and a15 = a + (15 – 1)d = 62
⇒ a + 14d = 62 ………2
From equation 2 – 1, we get
62 – 22 = 14d – 4d
⇒ 10d = 40
⇒ d = 4
From equation 1, we get
a + 4 × 4 = 22
⇒ a + 16 = 22
⇒ a = 6
Now,
a100 = 6 + 4(100 – 1 )
⇒ a100 = 6 + 4 × 99
⇒ a100 = 6 + 396
⇒ a100 = 402

Question 9.
Suppose a, b, c are in A.P. and a², b², c² are in G.P. If a < b < c and a + b + c = 3/2, then the value of a is
(a) 1/2√2
(b) 1/2√3
(c) 1/2 – 1/√3
(d) 1/2 – 1/√2

Given, a, b, c are in AP
⇒ 2b = a + c
⇒ b = (a + c)/2 ………….. 1
Again given, a², b², c² are in GP then b4 = a² c²
⇒ b² = ± ac ………… 2
Using 1 in a + b + c = 3/2, we get
3b = 3/2
⇒ b = 1/2
hence a + c = 1
and ac = ± 1/4
So a & c are roots of either x2 −x + 1/4 = 0 or x² − x − 1/4 = 0
The first has equal roots of x = 1/2 and second gives x= (1 ± √2)/2 for a and c
Since a < c,
we must have a = (1−√2)/2
⇒ a – 1/2 – √2/2
⇒ a – 1/2 – √2/(√2×√2)
⇒ a – 1/2 – 1/√2

Question 10.
If the positive numbers a, b, c, d are in A.P., then abc, abd, acd, bcd are
(a) not in A.P. / G.P. / H. P.
(b) in A.P.
(c) in G.P.
(d) in H.P.

Given, the positive numbers a, b, c, d are in A.P.
⇒ 1/a, 1/b, 1/c, 1/d are in H.P.
⇒ 1/d, 1/c, 1/b, 1/a are in H.P.
Now, Multiply by abcd, we get
abcd/d, abcd/c, abcd/b, abcd/a are in H.P.
⇒ abc, abd, acd, bcd are in H.P.

Question 11.
Let Tr be the rth term of an A.P. whose first term is a and the common difference is d. If for some positive integers m, n, m ≠ n, T m = 1/n and T n = 1/m then (a-d) equals to
(a) 0
(b) 1
(c) 1/mn
(d) 1/m + 1/n

Given the first term is a and the common difference is d of the AP
Now, Tm = 1/n
⇒ a + (m – 1)d = 1/n ………… 1
and Tn = 1/m
⇒ a + (n – 1)d = 1/m ………. 2
From equation 2 – 1, we get
(m – 1)d – (n – 1)d = 1/n – 1/m
⇒ (m – n)d = (m – n)/mn
⇒ d = 1/mn
From equation 1, we get
a + (m – 1)/mn = 1/n
⇒ a = 1/n – (m – 1)/mn
⇒ a = {m – (m – 1)}/mn
⇒ a = {m – m + 1)}/mn
⇒ a = 1/mn
Now, a – d = 1/mn – 1/mn
⇒ a – d = 0

Question 12.
In the sequence obtained by omitting the perfect squares from the sequence of natural numbers, then 2011th term is
(a) 2024
(b) 2036
(c) 2048
(d) 2055

Before 2024, there are 44 squares,
So, 1980th term is 2024
Hence, 2011th term is 2055

Question 13.
If the first term minus third term of a G.P. = 768 and the third term minus seventh term of the same G.P. = 240, then the product of first 21 terms =
(a) 1
(b) 2
(c) 3
(d) 4

Let first term = a
and common ratio = r
Given, a – ar² = 768
⇒ a(1 – r²) = 768
and ar² – ar6 = 240
⇒ ar² (1 – r4) = 240
Dividing the above 2 equations, we get
ar² (1 – r4)/a(1 – r²) = 240/768
⇒ {ar² (1 – r²) × (1 + r²)}/a(1 – r²) = 240/768
⇒ 1 + r² = 0.3125
⇒ r² = 0.25
⇒ r² = 25/100
⇒ r² = √(1/4)
⇒ r = ± 1/2
Now, a(1 – r²) = 768
⇒ a(1 – 1/4 ) = 768
⇒ 3a/4 = 768
⇒ 3a = 4 × 768
⇒ a = (4 × 768)/3
⇒ a = 4 × 256
⇒ a = 1024
⇒ a = 210
Now product of first 21 terms = (a² × r20)10 × a × r10
= a21 × r210
= (210)21 × (1/2)210
= 2210 /2210
= 1

Question 14.
If the sum of the first 2n terms of the A.P. 2, 5, 8, ….., is equal to the sum of the first n terms of the A.P. 57, 59, 61, ….., then n equals
(a) 10
(b) 12
(c) 11
(d) 13

Given, the sum of the first 2n terms of the A.P. 2, 5, 8, ….. = the sum of the first n terms of the A.P. 57, 59, 61, ….
⇒ (2n/2) × {2 × 2 + (2n – 1)3} = (n/2) × {2 × 57 + (n – 1)2}
⇒ n × {4 + 6n – 3} = (n/2) × {114 + 2n – 2}
⇒ 6n + 1 = {2n + 112}/2
⇒ 6n + 1 = n + 56
⇒ 6n – n = 56 – 1
⇒ 5n = 55
⇒ n = 55/5
⇒ n = 11

Question 15.
If a, b, c are in GP then log aⁿ, log bⁿ, log cⁿ are in
(a) AP
(b) GP
(c) Either in AP or in GP
(d) Neither in AP nor in GP

Given, a, b, c are in GP
⇒ b² = ac
⇒ (b²)ⁿ = (ac)ⁿ
⇒ (b2 )ⁿ= aⁿ × cⁿ
⇒ log (b²)ⁿ = log(an × cn )
⇒ log b²ⁿ = log aⁿ + log cⁿ
⇒ log (bⁿ)² = log aⁿ + log cⁿ
⇒ 2 × log bⁿ = log aⁿ + log cⁿ
⇒ log aⁿ, log bⁿ, log cⁿ are in AP

Question 16.
If the nth term of an AP is 3n – 4, the 10th term of AP is
(a) 12
(b) 22
(c) 28
(d) 30

Given, an = 3n – 2
Put n = 10, we get
a10 = 3 × 10 – 2
⇒ a10 = 30 – 2
⇒ a10 = 28
So, the 10th term of AP is 28

Question 17.
If the third term of an A.P. is 7 and its 7 th term is 2 more than three times of its third term, then the sum of its first 20 terms is
(a) 228
(b) 74
(c) 740
(d) 1090

Let a is the first term and d is the common difference of AP
Given the third term of an A.P. is 7 and its 7th term is 2 more than three times of its third term
⇒ a + 2d = 7 ………….. 1
and
3(a + 2d) + 2 = a + 6d
⇒ 3 × 7 + 2 = a + 6d
⇒ 21 + 2 = a + 6d
⇒ a + 6d = 23 ………….. 2
From equation 1 – 2, we get
4d = 16
⇒ d = 16/4
⇒ d = 4
From equation 1, we get
a + 2 × 4 = 7
⇒ a + 8 = 7
⇒ a = -1
Now, the sum of its first 20 terms
= (20/2) × {2 × (-1) + (20-1) × 4}
= 10 × {-2 + 19 × 4)}
= 10 × {-2 + 76)}
= 10 × 74
= 740

Question 18.
If a, b, c are in AP then
(a) b = a + c
(b) 2b = a + c
(c) b² = a + c
(d) 2b² = a + c

Answer: (b) 2b = a + c
Given, a, b, c are in AP
⇒ b – a = c – b
⇒ b + b = a + c
⇒ 2b = a + c

Question 19.
If 1/(b + c), 1/(c + a), 1/(a + b) are in AP then
(a) a, b, c are in AP
(b) a², b², c² are in AP
(c) 1/1, 1/b, 1/c are in AP
(d) None of these

Answer: (b) a², b², c² are in AP
Given, 1/(b + c), 1/(c + a), 1/(a + b)
⇒ 2/(c + a) = 1/(b + c) + 1/(a + b)
⇒ 2b2 = a² + c²
⇒ a², b², c² are in AP

Question 20.
3, 5, 7, 9, …….. is an example of
(a) Geometric Series
(b) Arithmetic Series
(c) Rational Exponent
(d) Logarithm