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.

## Sequences and Series Class 11 MCQs Questions with Answers

Question 1.

If a, b, c are in G.P., then the equations ax² + 2bx + c = 0 and dx² + 2ex + f = 0 have a common root if d/a, e/b, f/c are in

(a) AP

(b) GP

(c) HP

(d) none of these

## Answer

Answer: (a) AP

Hint:

Given a, b, c are in GP

⇒ b² = ac

⇒ b² – ac = 0

So, ax² + 2bx + c = 0 have equal roots.

Now D = 4b² – 4ac

and the root is -2b/2a = -b/a

So -b/a is the common root.

Now,

dx² + 2ex + f = 0

⇒ d(-b/a)² + 2e×(-b/a) + f = 0

⇒ db2 /a² – 2be/a + f = 0

⇒ d×ac /a² – 2be/a + f = 0

⇒ dc/a – 2be/a + f = 0

⇒ d/a – 2be/ac + f/c = 0

⇒ d/a + f/c = 2be/ac

⇒ d/a + f/c = 2be/b²

⇒ d/a + f/c = 2e/b

⇒ d/a, e/b, f/c are in AP

Question 2.

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

Answer: (b) 2b = a + c

Hint:

Given, a, b, c are in AP

⇒ b – a = c – b

⇒ b + b = a + c

⇒ 2b = a + c

Question 3:

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 + √3

(b) 2 – √3

(c) 2 ± √3

(d) None of these

## Answer

Answer: (a) 2 + √3

Hint:

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 ± √3

⇒ r = 2 + √3 {Since r > 1}

Question 4:

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

## Answer

Answer: (a) n/(n+1)

Hint:

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 5:

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

Answer: (b) a², b², c² are in AP

Hint:

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 6:

The sum of series 1/2! + 1/4! + 1/6! + ….. is

(a) e² – 1 / 2

(b) (e – 1)² /2 e

(c) e² – 1 / 2 e

(d) e² – 2 / e

## Answer

Answer: (b) (e – 1)² /2 e

Hint:

We know that,

e^{x} = 1 + x/1! + x² /2! + x³ /3! + x^{4} /4! + ………..

Now,

e^{1} = 1 + 1/1! + 1/2! + 1/3! + 1/4! + ………..

e^{-1} = 1 – 1/1! + 1/2! – 1/3! + 1/4! + ………..

e^{1} + e^{-1} = 2(1 + 1/2! + 1/4! + ………..)

⇒ e + 1/e = 2(1 + 1/2! + 1/4! + ………..)

⇒ (e² + 1)/e = 2(1 + 1/2! + 1/4! + ………..)

⇒ (e² + 1)/2e = 1 + 1/2! + 1/4! + ………..

⇒ (e² + 1)/2e – 1 = 1/2! + 1/4! + ………..

⇒ (e² + 1 – 2e)/2e = 1/2! + 1/4! + ………..

⇒ (e – 1)² /2e = 1/2! + 1/4! + ………..

Question 7:

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

(a) 4^{3}

(b) 4^{5}

(c) 4^{4}

(d) none of these

## Answer

Answer: (b) 4^{5}

Hint:

here it is given that T_{3} = 4.

⇒ ar² = 4

Now product of first five terms = a.ar.ar².ar³.ar^{4}

= a^{5}r^{10}

= (ar^{2})^{5}

= 4^{5}

Question 8:

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 Tm = 1/n and Tn = 1/m, then Tm n equals

(a) 1/m n

(b) 1/m + 1/n

(c) 1

(d) 0

## Answer

Answer: (c) 1

Hint:

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

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

## Answer

Answer: (c) 6

Hint:

Let a and b are two numbers such that

a + b = 13/6

Let A_{1}, A_{2}, A_{3}, ………A_{2n} be 2n arithmetic means between a and b

Then, A_{1} + A_{2} + A_{3} + ………+ A_{2n} = 2n{(n + 1)/2}

⇒ n(a + b) = 13n/6

Given that A_{1} + A_{2} + A_{3} + ………+ A_{2n} = 2n + 1

⇒ 13n/6 = 2n + 1

⇒ n = 6

Question 10.

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.

## Answer

Answer: (c) H.P.

Hint:

Given, equation is

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

Question 11.

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

Answer: (b) a², b², c² are in AP

Hint:

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 12.

The 35th partial sum of the arithmetic sequence with terms a_{n} = n/2 + 1

(a) 240

(b) 280

(c) 330

(d) 350

## Answer

Answer: (d) 350

Hint:

The 35th partial sum of this sequence is the sum of the first thirty-five terms.

The first few terms of the sequence are:

a_{1} = 1/2 + 1 = 3/2

a_{2} = 2/2 + 1 = 2

a_{3} = 3/2 + 1 = 5/2

Here common difference d = 2 – 3/2 = 1/2

Now, a_{35} = a_{1} + (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

Question 13.

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

## Answer

Answer: (c) 6

Hint:

Let a and b are two numbers such that

a + b = 13/6

Let A_{1}, A_{2}, A_{3}, ………A_{2n} be 2n arithmetic means between a and b

Then, A_{1} + A_{2} + A_{3} + ………+ A_{2n} = 2n{(n + 1)/2}

⇒ n(a + b) = 13n/6

Given that A_{1} + A_{2} + A_{3} + ………+ A_{2n} = 2n + 1

⇒ 13n/6 = 2n + 1

⇒ n = 6

Question 14.

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

## Answer

Answer: (c) 3

Hint:

Let first term of the GP is a and common ratio is r.

3rd term = ar²

5th term = ar^{4}

Now

⇒ ar² + ar^{4} = 90

⇒ a(r² + r^{4}) = 90

⇒ r² + r^{4} = 90

⇒ r² ×(r² + 1) = 90

⇒ r²(r² + 1) = 3² ×(3² + 1)

⇒ r = 3

So the common ratio is 3

Question 15.

The sum of AP 2, 5, 8, …..up to 50 terms is

(a) 3557

(b) 3775

(c) 3757

(d) 3575

## Answer

Answer: (b) 3775

Hint:

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 16.

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

## Answer

Answer: (b) 31/48

Hint:

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 17.

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

## Answer

Answer: (a) n/(n+1)

Hint:

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 18.

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

## Answer

Answer: (c) 740

Hint:

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 19.

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

## Answer

Answer: (c) 11

Hint:

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 20.

If a is the A.M. of b and c and G_{1} and G_{2} are two GM between them then the sum of their cubes is

(a) abc

(b) 2abc

(c) 3abc

(d) 4abc

## Answer

Answer: (b) 2abc

Hint:

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

⇒ a = (b + c)

⇒ 2a = b + c ………… 1

Again, given G_{1} and G_{1} are two GM between b and c,

⇒ b, G_{1}, G_{2}, c are in the GP having common ration r, then

⇒ r = (c/b)^{1/(2+1)} = (c/b)^{1/3}

Now,

G_{1} = br = b×(c/b)^{1/3}

and G_{1} = br = b×(c/b)^{2/3}

Now,

(G_{1})³ + (G_{2})3 = b³ ×(c/b) + b³ ×(c/b)²

⇒ (G_{1})³ + (G_{2})³ = b³ ×(c/b)×( 1 + c/b)

⇒ (G_{1})³ + (G_{2})³ = b³ ×(c/b)×( b + c)/b

⇒ (G_{1})³ + (G_{2})³ = b² ×c×( b + c)/b

⇒ (G_{1})³ + (G_{2})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

(G_{1})³ + (G_{2})³ = b² × c × (2a/b)

⇒ (G_{1})³ + (G_{2})³ = b × c × 2a

⇒ (G_{1})³ + (G_{2})³ = 2abc

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