Class 11 Maths Chapter 9 Sequences and Series MCQs

Class 11 Maths Chapter 9 Sequences and Series MCQs are available online for students to achieve good exam outcomes. The multiple-choice questions are based on the CBSE syllabus and NCERT guidelines (2022-2023). These multiple-choice questions have complete answers and explanations. To obtain all chapter-by-chapter MCQs for Class 11 Maths, click here.

MCQs on Class 11 Maths Chapter 9 Sequences and Series

Go through the MCQs for Class 11 Maths Chapter 9 sequences and series. For each MCQ, there are four options, but only one is correct.

Download PDF – Chapter 9 Sequence and Series MCQs

Students should choose the appropriate option and check their answers against the solutions on our website. Take a look at the important questions for class 11 Maths as well.

1) If a, 4, b are in Arithmetic Progression; a, 2, b are in Geometric Progression; then a, 1, b are in

  1. A.P
  2. G.P
  3. H.P
  4. None of these

Answer: (c) H.P

Explanation:

Given that a, 4, b are in A.P

Hence, 4-a = b-4

a+b = 8 …(1)

Also, given that a, 2, b are in G.P.

Hence, 2/a = b/2

So, ab = 4…(2)

If a, 1, b are in H.P, then 1 = 2(ab)/(a+b) ..(3)

Now substitute (1) and (2) in (3)

1 = 2(4) /(8)

1 = 8/8

1=1.

Therefore, a, 1, b are in H.P.

2) If “a” is the first term and “r” is the common ratio, then the nth term of a G.P is:

  1. arn
  2. arn-1
  3. (ar)n-1
  4. None of these

Answer: (b) arn-1

Explanation:

If “a” is the first term and “r” is the common ratio, the terms of infinite G.P are written as a, ar, ar2, ar3, ar4, …arn-1.

Hence, the nth term of a G.P is arn-1.

Therefore, option (b) is the correct answer.

3) If a, b, c are in arithmetic progression, then

  1. b = a+c
  2. 2b = a+c
  3. b2 = a+c
  4. 2b2 = a+c

Answer: (b) 2b = a+c

Explanation:

Given that a, b, c are in arithmetic progression.

So, the common difference is b-a = c-b

Rearranging the same terms, we get

b+b = c+a

2b = a+c.

Hence, if a, b, c are in A.P, then 2b = a+c.

4) The sum of arithmetic progression 2, 5, 8, …, up to 50 terms is

  1. 3775
  2. 3557
  3. 3757
  4. 3575

Answer: (a) 3775

Explanation:

Given A.P. = 2, 5, 8, …

We know that the sum of n terms of an A.P is Sn = (n/2)[2a+(n-1)d]

Here, a = 2, d = 3 and n=50.

Now, substitute the values in the formula, we get

S50 = (50/2)[2(2)+(50-1)(3)]

S50 = 25[4+(49)(3)]

S50 = 25[4+147]

S50 = 25(151)

S50 = 3775.

Hence, the sum of A.P 2, 5, 8, …up to 50 terms is 3775.

5) The 3rd term of G.P is 4. Then the product of the first 5 terms is:

  1. 43
  2. 44
  3. 45
  4. None of these

Answer: (c) 45

Explanation:

We know that the terms of infinite G.P are written as a, ar, ar2, ar3, ar4, …arn-1.

Hence, the 3rd term, (i.e) ar2= 4

Thus, the product of the first 5 terms = (a)(ar)(ar2)(ar3)(ar4)

= a5r10

= (ar2)5

Now, substitute ar2 = 4 in the above form, we get

Product of the first 5 terms = (4)5 = 45.

Hence, option (c) 45 is the correct answer.

6) Which of the following is an example of a geometric sequence?

  1. 1, 2, 3, 4
  2. 1, 2, 4, 8
  3. 3, 5, 7, 9
  4. 9, 20, 21, 28

Answer: (b) 1, 2, 4, 8

Explanation:

Among the options given, option (b) 1, 2, 4, 8 is an example of a geometric sequence.

We know that in a geometric sequence each term is found by multiplying the previous term by a constant.

In option (b) 1, 2, 4, 8, each term is found by multiplying 2 to the previous term. Here, the common ratio is 2.

7) The next term of the given sequence 1, 5, 14, 30, 55, … is

  1. 80
  2. 90
  3. 91
  4. 96

Answer: (c) 91

Explanation: The next term in the sequence 1, 5, 14, 30, 55, … is 91.

Ist term = 12 = 1

2nd term = 12+22 = 1+4 = 5

3rd term = 12+22+32 = 1+4+9 = 14

4th term = 12+22+32+42 = 1+4+9+16 = 30

5th term = 12+22+32+42+52 = 1+4+9+16+25 = 55

6th term = 12+22+32+42+52+62 = 1+4+9+16+25+36 = 91.

Thus, option (c) is the correct answer.

8) If the nth term of an arithmetic progression is 3n-4, then the 10th term of an A.P is

  1. 10
  2. 12
  3. 22
  4. 26

Answer: (d) 26

Explanation:

Given that the nth term of A.P = 3n-4.

To find the 10th term of A.P, substitute n = 10

Therefore, 10th term of A.P = 3(10) -4 = 30-4 = 26.

9) 3, 5, 7, 9 is an example of

  1. Arithmetic sequence
  2. Geometric sequence
  3. Harmonic sequence
  4. Fibonacci sequence

Answer: (a) Arithmetic sequence

Explanation: 3, 5, 7, 9 is an example of an arithmetic sequence. In this sequence 3, 5, 7, 9, the difference between each term is 2.

(i.e) 5-3 = 2, 7-5 = 2, 9-7 = 2.

Hence 3, 5, 7, 9 is an arithmetic sequence.

10) The first term of a G.P is 1. The sum of the 3rd and 5th terms is 90. Then the common ratio is:

  1. 1
  2. 2
  3. 3
  4. 4

Answer: (c) 3

Explanation:

Given that first term of G.P, a = 1.

The sum of the 3rd and 5th term = 90

(i.e) ar2+ar4 = 90

Substitute a = 1,

⇒ r2+r4 = 90

⇒ r4 + r2 – 90 = 0

⇒ r4 + 10r2 – 9r2 – 90 = 0

Now, factorize the above equation,

⇒ r2 (r2+10) – 9 (r2+10) = 0

⇒ (r2-9)(r2+10) = 0

⇒ r2 = 9 or r2 = -10

Here, r2= -10 is not possible, as the square of a number cannot be negative.

So, r2 = 9

r = 3 or r= -3

Therefore, option (c) 3 is the correct answer.

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