Important Class 11 Maths MCQ Questions of Sequences and Series with Answers are given here. Sequences and series have a few applications in our everyday life. The significant MCQ Questions for class 11 Maths cover every one of the topics and subtopics in the NCERT.

Go through the significant MCQ Questions gave at “Sarthaks eConnect” and accomplish astounding marks in the yearly assessment.  It’s based on latest syllabus and exam pattern. The answers of the questions are given in a detailed explanation. so, students can understand them in a superior manner.

Practice MCQ Questions for class 11 Maths Chapter-Wise

1. If a, b, c are in AP then

(a) b = a + c
(b) 2b = a + c
(c) b² = a + c
(d) 2b² = a + c

2. Three numbers form an increasing GP. If the middle term is doubled, then the new numbers are in Ap. The common ratio of GP is

(a) 2 + \(\sqrt3\)
(b) 2 – \(\sqrt3\)
(c) 2 ± \(\sqrt3\)
(d) None of these

3. 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

4.  The sum of the series 1/2! − 1/3! + 1/4! −... up to infinity is

(a) e^-2
(b) e^-1
(c) e^-1/2
(d) e^1/2

5. The third term of a geometric progression is 4. The product of the first five terms is

(a) 4³
(b) 4⁵
(c) 4⁴
(d) none of these

6. Let Tr be the r^th term of an A.P., for r = 1, 2, 3, … If for some positive integers m, n, we have T_m = 1/n and T_n = 1/m, then T_mn equals

(a) 1/mn
(b) 1/m + 1/n
(c) 1
(d) 0

7. 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

8. If the sum of the roots of the quadratic equation ax² + bx + c = 0 is equal to the sum of the squares of their reciprocals, then a/c, b/a, c/b are in

(a) A.P
(b) G.P
(c) H.P
(d) A.G.P

9. 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

10. The 35^th partial sum of the arithmetic sequence with terms an = n/2 + 1

(a) 240
(b) 280
(c) 330
(d) 350

11. 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

12. 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

13. 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

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

15. 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

16. 3, 5, 7, 9, …….. is an example of

(a) Geometric Series
(b) Arithmetic Series
(c) Rational Exponent
(d) Logarithm

17.  2, 3, 5, 7, 11, ?, 17

(a) 12
(b) 13
(c) 14
(d) 15

18. If the positive numbers a,b,c,d are in AP. Then, abc,abd,acd,bcd are

(a) not in AP/GP/HP
(b) in AP
(c) in GP
(d) in HP

19. The minimum value of expression 3^x+3^1−x,x ∈ R, is

(a) 0
(b) 1/3
(c) 3
(d) \(2\sqrt3\)

20. The sum of odd integers from 1 to 2001 is

(a) 10201
(b) 102001
(c) 100201
(d) 1002001

Answer:

1. Answer: (b) 2b = a + c

Explanation: Given, a, b, c are in AP

⇒ b – a = c – b

⇒ b + b = a + c

⇒ 2b = a + c

2. Answer: (a) 2+ \(\sqrt3\)

Explanation: Let the three numbers be a/r, a, ar

Since the numbers form an increasing GP, So r > 1

Now, it is given that a/r, 2a, ar are in AP

⇒ 4a = a/r + ar

⇒ r² – 4r + 1 = 0

⇒ r = 2 ± \(\sqrt3\)

⇒ r = 2 + \(\sqrt3\) {Since r > 1}

3. Answer: (a) n/(n+1)

Explanation: 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)

4. Answer: (b) e^-1

Explanation: e^−x = 1−x + x²/ 2! − x³/3! + x⁴/4!−... .....

put x = 1

1/2! − 1/3! + 1/4! .....= e^-1

5. Answer: (b) 4⁵

Explanation: here it is given that T3 = 4.

⇒ ar² = 4

Now product of first five terms = a.ar.ar².ar³.ar4

= a⁵r¹⁰

= (ar²)⁵

= 4⁵

6. Answer: (c) 1

Explanation: Let first term is a and the common difference is d of the AP

Now, T_m = 1/n

⇒ a + (m-1)d = 1/n ………… 1

and T_n = 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, T_mn = 1/mn + (mn-1)/mn

⇒ T_mn = 1/mn + 1 – 1/mn

⇒ T_mn = 1

7.  Answer: (c) 6

Explanation: Let a and b are two numbers such that

a + b = 13/6

Let A₁, A₂, A₃, ………A_2n be 2n arithmetic means between a and b

Then, A₁ + A₂ + A₃ + ………+ A_2n = 2n{(n + 1)/2}

⇒ n(a + b) = 13n/6

Given that A₁ + A₂ + A₃ + ………+ A_2n = 2n + 1

⇒ 13n/6 = 2n + 1

⇒ n = 6

8. Answer: (c) H.P

Explanation: (c) H.P

ax² + bx + c = 0

Let p and q are the roots of this equation.

Now p+q = -b/a

and pq = c/a

Given that

p + q = 1/p² + 1/q²

⇒ p + q = (p² + q²)/(p² ×q²)

⇒ p + q = {(p + q)² – 2pq}/(pq)²

⇒ -b/a = {(-b/a)² – 2c/a}/(c/a)²

⇒ (-b/a)×(c/a)² = {b²/a² – 2c/a}

⇒ -bc²/a³ = {b² – 2ca}/a²

⇒ -bc²/a = b² – 2ca

Divide by bc on both side, we get

⇒ -c /a = b/c – 2a/b

⇒ 2a/b = b/c + c/a

⇒ b/c, a/b, c/a are in AP

⇒ c/a, a/b, b/c are in AP

⇒ 1/(c/a), 1/(a/b), 1/(b/c) are in HP

⇒ a/c, b/a, c/b are in HP

9. Answer: (b) a², b², c² are in AP

Explanation: 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

10. Answer: (d) 350

Explanation: 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

11. Answer: (c) 3

Explanation: 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

12. Answer: (b) 31/48

Explanation: 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

13. Answer: (c) 740

Explanation: 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

14. Answer: (c) 11

Explanation: 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

15. Answer: (b) 2abc

Explanation:  Given, a is the A.M. of b and c

⇒ a = (b + c)

⇒ 2a = b + c ………… 1

Again, given G₁ and G₂ 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 = b × c × (2a/b)

⇒ (G₁)³ + (G₂)² = b × c × 2a

⇒ (G₁)³ + (G₂)³ = 2abc

16. Answer: (b) Arithmetic Series

Explanation: This is an arithmetic sequence since there is a common difference between each term.

17. Answer: 13

Explanation: Clearly, the given series consists of prime numbers starting from 2. So, the missing term is the prime number after 11, which is 13.

18. Answer: (d) in HP

Explanation: Since, a,b,c,d are in AP.

⇒ a/abcd,b/abcd,c/abcd,d/abcd are in AP

⇒1/bcd,1/cda,1/abd,1/abc are in AP

⇒ bcd,cda,abd,abc are in HP

⇒ abc,abd,cda,bcd are in HP

19. Answer: (d) \(2\sqrt3\)

Explanation: We know that A.M.≥G.M. for positive numbers.

\(\frac{3^x+3^{1-x}}{2}\) \(\geq\sqrt{3^x.3^{1-x}}\)

⇒3^x + 3^1−x ≥ \(2\sqrt3\)

20. Answer: (d) 1002001 

Explanation: The odd integers from 1 to 2001 are 1, 3, 5, …1999, 2001.

This sequence forms an A.P.

Here, first term, a = 1

Common difference, d = 2

Here,

a+(n−1)d = 2001

=> 1+(n−1)(2) = 2001

=> 2n−2 = 2000

=> n = 1001

S_n = n/2[2a+(n−1)d]

∴ S_n = 1001/2[2×1+(1001−1)×2]

=1001/2 [2+1000×2]

=1001/2×2002

=1001×1001

=1002001

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