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## Sequences and Series Class 11 MCQs Questions with Answers

Students are advised to solve the Sequences and Series Multiple Choice Questions of Class 11 Maths to know different concepts. Practicing the MCQ Questions on Sequences and Series Class 11 with answers will boost your confidence thereby helping you score well in the exam.

Explore numerous MCQ Questions of Sequences and Series Class 11 with answers provided with detailed solutions by looking below.

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

## Answer

Answer: (a) 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

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

## Answer

Answer: (c) 3

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 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 × r^{n-1}

(d) None of these

## Answer

Answer: (c) 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}

Question 4.

The sum of odd integers from 1 to 2001 is

(a) 10201

(b) 102001

(c) 100201

(d) 1002001

## Answer

Answer: (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 x^{b-c} × y^{c-a} × z^{a-b} is

(a) 0

(b) 1

(c) -1

(d) None of these

## Answer

Answer: (b) 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

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

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

## Answer

Answer: (c) 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

## Answer

Answer: (c) 402

Let ais the first term and d is the common difference of the AP

Given,

a_{5} = 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

## Answer

Answer: (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 b^{4} = 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.

## Answer

Answer: (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

## Answer

Answer: (a) 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

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

## Answer

Answer: (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

## Answer

Answer: (a) 1

Let first term = a

and common ratio = r

Given, a – ar² = 768

⇒ a(1 – r²) = 768

and ar² – ar^{6} = 240

⇒ ar² (1 – r^{4}) = 240

Dividing the above 2 equations, we get

ar² (1 – r^{4})/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 = 2^{10}

Now product of first 21 terms = (a² × r^{20})^{10} × a × r^{10}

= a^{21} × r^{210}

= (2^{10})^{21} × (1/2)^{210}

= 2^{210} /2^{210}

= 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

## Answer

Answer: (c) 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

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

## Answer

Answer: (a) 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

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

## Answer

Answer: (c) 28

Given, a_{n} = 3n – 2

Put n = 10, we get

a_{10} = 3 × 10 – 2

⇒ a_{10} = 30 – 2

⇒ a_{10} = 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

## Answer

Answer: (c) 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

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

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

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

## Answer

Answer: (b) Arithmetic Series

3, 5, 7, 9, …….. is an example of Arithmetic Series.

Here common difference = 5 – 3 = 7 – 5 = 9 – 7 = 2

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