CBSE Class 11 Mathematics Chapter 9 Sequences and Series Multiple Choice Questions with Answers. MCQ Class 11 Mathematics Sequences and Series with Answers was Prepared Based on Latest Exam Pattern. Students can solve NCERT Class 11 Mathematics Sequences and Series MCQs with Answers to know their preparation level.
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MCQ Class 11 Mathematics Sequences and Series with Answers - Set - 2
Question 1:
If a is the A.M. of b and c and G1 and G2 are two GM between them then the sum of their cubes is
(a) abc
(b) 2abc
(c) 3abc
(d) 4abc
Correct Answer – (B)
Given, a is the A.M. of b and c
⇒ a = (b + c)
⇒ 2a = b + c ………… 1
Again, given G1 and G1 are two GM between b and c,
⇒ b, G1, G2, c are in the GP having common ration r, then
⇒ r = (c/b)1/(2+1) = (c/b)1/3
Now,
G1 = br = b×(c/b)1/3
and G1 = br = b×(c/b)2/3
Now,
(G1)³ + (G2)3 = b³ ×(c/b) + b³ ×(c/b)²
⇒ (G1)³ + (G2)³ = b³ ×(c/b)×( 1 + c/b)
⇒ (G1)³ + (G2)³ = b³ ×(c/b)×( b + c)/b
⇒ (G1)³ + (G2)³ = b² ×c×( b + c)/b
⇒ (G1)³ + (G2)³ = b² ×c×( b + c)/b ………….. 2
From equation 1
2a = b + c
⇒ 2a/b = (b + c)/b
Put value of(b + c)/b in eqaution 2, we get
(G1)³ + (G2)³ = b² × c × (2a/b)
⇒ (G1)³ + (G2)³ = b × c × 2a
⇒ (G1)³ + (G2)³ = 2abc
Question 2 :
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
Correct Answer – (C)
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 3 :
If 2/3, k, 5/8 are in AP then the value of k is
(a) 31/24
(b) 31/48
(c) 24/31
(d) 48/31
Correct Answer – (B)
Given, 2/3, k, 5/8 are in AP
⇒ 2k = 2/3 + 5/8
⇒ 2k = 31/24
⇒ k = 31/48
So, the value of k is 31/48
Question 4 :
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
Correct Answer – (C)
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 5 :
The 35th partial sum of the arithmetic sequence with terms an = n/2 + 1
(a) 240
(b) 280
(c) 330
(d) 350
Correct Answer – (D)
The 35th partial sum of this sequence is the sum of the first thirty-five terms.
The first few terms of the sequence are:
a1 = 1/2 + 1 = 3/2
a2 = 2/2 + 1 = 2
a3 = 3/2 + 1 = 5/2
Here common difference d = 2 – 3/2 = 1/2
Now, a35 = a1 + (35 – 1)d = 3/2 + 34 ×(1/2) = 17/2
Now, the sum = (35/2) × (3/2 + 37/2)
= (35/2) × (40/2)
= (35/2) × 20
= 35 × 10
= 350
MCQ Class 11 Mathematics Sequences and Series with Answers
Question 6 :
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
Correct Answer – (C)
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 7 :
The sum of n terms of the series (1/1.2) + (1/2.3) + (1/3.4) + …… is
(a) n/(n+1)
(b) 1/(n+1)
(c) 1/n
(d) None of these
Correct Answer – (A)
Given series is:
S = (1/1·2) + (1/2·3) + (1/3·4) – ……………….1/n.(n+1)
⇒ S = (1 – 1/2) + (1/2 – 1/3) + (1/3 – 1.4) -………(1/n – 1/(n+1))
⇒ S = 1 – 1/2 + 1/2 – 1/3 + 1/3 – 1/4 – ……….. 1/n – 1/(n+1)
⇒ S = 1 – 1/(n+1)
⇒ S = (n + 1 – 1)/(n+1)
⇒ S = n/(n+1)
Question 8 :
The sum of AP 2, 5, 8, …..up to 50 terms is
(a) 3557
(b) 3775
(c) 3757
(d) 3575
Correct Answer – (B)
Given, AP is 2, 5, 8, …..up to 50
Now, first term a = 2
common difference d = 5 – 2 = 3
Number of terms = 50
Now, Sum = (n/2)×{2a + (n – 1)d}
= (50/2)×{2×2 + (50 – 1)3}
= 25×{4 + 49×3}
= 25×(4 + 147)
= 25 × 151
= 3775
Question 9 :
The sum of two numbers is 13/6 An even number of arithmetic means are being inserted between them and their sum exceeds their number by 1. Then the number of means inserted is
(a) 2
(b) 4
(c) 6
(d) 8
Correct Answer – (C)
Let a and b are two numbers such that
a + b = 13/6
Let A1, A2, A3, ………A2n be 2n arithmetic means between a and b
Then, A1 + A2 + A3 + ………+ A2n = 2n{(n + 1)/2}
⇒ n(a + b) = 13n/6
Given that A1 + A2 + A3 + ………+ A2n = 2n + 1
⇒ 13n/6 = 2n + 1
⇒ n = 6
Question 10 :
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
Correct Answer – (B)
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
- NCERT Solutions Class 11 Mathematics Sequences and Series with Answers : Exercise 9.1
- NCERT Solutions Class 11 Mathematics Sequences and Series with Answers : Exercise 9.2
- NCERT Solutions Class 11 Mathematics Sequences and Series with Answers : Exercise 9.3
- NCERT Solutions Class 11 Mathematics Sequences and Series with Answers : Exercise 9.4
- NCERT Solutions Class 11 Mathematics Sequences and Series with Answers : Exercise 9 Misc
- NCERT Solutions Class 11 Mathematics Textbook download
