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

Sequences and Series Class 11 Maths MCQs Questions with Answers

Check the below NCERT MCQ Questions for Class 11 Maths Chapter 9 Sequences and Series with Answers Pdf free download. MCQ Questions for Class 11 Maths with Answers were prepared based on the latest exam pattern. We have Provided Sequences and Series Class 11 Maths MCQs Questions with Answers to help students understand the concept very well.

Class 11 Maths Chapter 9 Quiz

Class 11 Maths Chapter 9 MCQ Online Test

You can refer to NCERT Solutions for Class 11 Maths Chapter 9 Sequences and Series to revise the concepts in the syllabus effectively and improve your chances of securing high marks in your board exams.

Sequences and Series Class 11 Maths MCQ online test

Q1.	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

Ans: 0
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

Q2.	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

Ans: 3
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

Q3.	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 × r^n-1
	D.None of these

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

Q4.	The sum of odd integers from 1 to 2001 is
	A.10201
	B.102001
	C.100201
	D.1002001

Ans: 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

Q5.	If a, b, c are in AP and x, y, z are in GP then the value of x^b-c × y^c-a × z^a-b is
	A.0
	B.1
	C.-1
	D.None of these

Ans: 1
Given, a, b, c are in AP
⇒ 2b = a + c ………. 1
and x, y, z are in GP
⇒ y² = xz ……….. 2
Now, x^b-c × y^c-a × z^a-b = x^b-c × (√xz)^c-a × z^a-b
= x^b-c × x^(c-a)/2 × z^(c-a)/2 × z^a-b
= x^b-c + x^(c-a)/2 × z^{(c-a)/2+ a -b}
= x^2b+(c+a) × z^(c+a)-2b
= x° × z°
= 1
So, the value of x^b-c × y^c-a × z^a-b is 1

Q6.	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

Ans: 1, 2, 4, 8
1, 2, 4, 8 is the example of geometric series
Here common ratio = 2/1 = 4/2 = 8/4 = 2

Q7.	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

Ans: 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)

Q8.	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

Ans: 402
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

Q9.	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

Ans: 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

Q10.	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.

Ans: 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.

Q11.	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

Q12.	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

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

Q13.	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

Ans: 1
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

Q14.	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

Ans: 11
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

Q15.	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

Ans: AP
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

Q16.	If the nth term of an AP is 3n – 4, the 10th term of AP is
	A.12
	B.22
	C.28
	D.30

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

Q17.	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

Ans: 740
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