# CBSE 10th Maths Exam 2021: Important MCQs from Chapter 5 Arithmetic Progression with Detailed Solutions

Class 10 Mathematics MCQs on Chapter 5 - Arithmetic Progression are provided here to prepare for the CBSE Board Exam 2021.

*CBSE Class 10 Maths MCQs Chapter 5 Arithmetic Progression*

Multiple Choice Questions (MCQs) on Arithmetic Progression of CBSE Class 10 Maths are provided here with answers and their explanation. These questions are good to prepare the objective type questions that will appear in the Class 10 Mathematics Board Question Paper 2021.

**Note - **Students must note that** 'Application in solving daily life problems based on sum to n terms'** **has been excluded from the revised CBSE syllabus of Class10 Maths**. So, students should prepare according to the new syllabus only.

**Check below the solved MCQs from Class 10 Maths Chapter 5 ****Arithmetic Progression:**

**1.** In an AP, if d = –4, n = 7, a_{n} = 4, then a is

(A) 6

(B) 7

(C) 20

(D) 28

**Answer:**** (D)**

**Explanation:**

For an A.P

a_{n} = a + (n – 1)d

4 = a + (7 – 1)( −4)

4 = a + 6(−4)

4 + 24 = a

a = 28

**2.** In an AP, if a = 3.5, d = 0, n = 101, then a_{n} will be

(A) 0

(B) 3.5

(C) 103.5

(D) 104.5

**Answer:**** (B)**

**Explanation:**

For an A.P

a_{n} = a + (n – 1)d

= 3.5 + (101 – 1) × 0

= 3.5

**3.** The first four terms of an AP, whose first term is –2 and the common difference is –2, are

(A) – 2, 0, 2, 4

(B) – 2, 4, – 8, 16

(C) – 2, – 4, – 6, – 8

(D) – 2, – 4, – 8, –16

**Answer:**** (C)**

**Explanation:**

Let the first four terms of an A.P are a, a+d, a+2d and a+3d

Given that the first termis −2 and difference is also −2, then the A.P would be:

– 2, (–2–2), [–2 + 2 (–2)], [–2 + 3(–2)]

= –2, –4, –6, –8

**4.** The famous mathematician associated with finding the sum of the first 100 natural numbers is

(A) Pythagoras

(B) Newton

(C) Gauss

(D) Euclid

**Answer:**** (C)**

**Explanation:**

Gauss is the famous mathematician associated with finding the sum of the first 100 natural Numbers.

(A) –20

(B) 20

(C) –30

(D) 30

**Answer: (B)**

**Explanation:**

**6.** The 21st term of the AP whose first two terms are –3 and 4 is

(A) 17

(B) 137

(C) 143

(D) –143

**Answer: (B)**

**Explanation:**

First two terms are –3 and 4

Therefore,

a = −3

a + d = 4

⇒ d = 4 − a

⇒ d = 4 + 3

⇒ d = 7

Thus,

a_{21} = a + (21 – 1)d

a_{21} = –3 + (20)7

a_{21} = 137

**Also Check: ****CBSE Class 10 Science Important MCQs: All Chapters**

**7.** If the 2nd term of an AP is 13 and the 5th term is 25, what is its 7th term?

(A) 30

(B) 33

(C) 37

(D) 38

**Answer: (B)**

**Explanation:**

Since

a_{2} = 13

a_{5} = 25

⇒ a + d = 13 ….(i)

⇒ a + 4d = 25 ….(ii)

Solving equations (i) and (ii), we get:

a = 9; d = 4

Therefore,

a_{7 }= 9 + 6 × 4

a_{7 }= 9 + 24

a_{7 }= 33

**8.** If the common difference of an AP is 5, then what is a_{18} – a_{13}?

(A) 5

(B) 20

(C) 25

(D) 30

**Answer:**** (C)**

**Explanation:**

Since, d = 5

a_{18} – a_{13} = a + 17d – a – 12d

= 5d

= 5 × 5

= 25

**9.** The sum of first 16 terms of the AP: 10, 6, 2,... is

(A) –320

(B) 320

(C) –352

(D) –400

**Answer: (A)**

Given A.P. is 10, 6, 2,...

**10.** The sum of first five multiples of 3 is

(A) 45

(B) 55

(C) 65

(D) 75

**Answer: (A)**

**Explanation:**

The first five multiples of 3 are 3, 6, 9, 12 and 15

**11.** The middle most term (s) of the AP:–11, –7, –3, ..., 49 is:

(A) 18, 20

(B) 19, 23

(C) 17, 21

(D) 23, 25

**Answer: (C)**

**Explanation:**

Here, a = −11

d = − 7 – (−11) = 4

And a_{n} = 49

We have,

a_{n} = a + (n – 1)d

⇒ 49 = −11 + (n – 1)4

⇒ 60 = (n – 1)4

⇒ n = 16

As n is an even number, there will be two middle terms which are16/2th and [(16/2)+1]th, i.e. the 8th term and the 9th term.

a_{8} = a + 7d = – 11 + 7 × 4 = 17

a_{9} = a + 8d = – 11 + 8 × 4 = 21

**12.** Two APs have the same common difference. The first term of one of these is –1 and that of the other is – 8. Then the difference between their 4th terms is

(A) –1

(B) – 8

(C) 7

(D) –9

**Answer: (C)**

**Explanation:**

The 4th term of first series is

a_{4} = a_{1} + 3d

The 4th term of another series is

a`_{4} = a_{2} + 3d

Now,

As, a_{1} = –1, a_{2} = –8

Therefore,

a_{4 }– a`_{4 }= (–1 + 3d) – (–8 + 3d)

a_{4 }– a`_{4 }= 7

**13.** If 7 times the 7th term of an AP is equal to 11 times its 11th term, then its 18th term will be

(A) 7

(B) 11

(C) 18

(D) 0

**Answer:**** (D)**

**Explanation:**

According to question

7(a + 6d) = 11(a + 10d)

⇒ 7a + 42d = 11a + 110d

⇒ 4a + 68d = 0

⇒ 4(a + 17d) = 0

⇒ a + 17d = 0

Therefore,

a_{18} = a + 17d

a_{18 }= 0

**14.** In an AP if a = 1, a_{n} = 20 and S_{n} = 399, then n is

(A) 19

(B) 21

(C) 38

(D) 42

**Answer: (C)**

**Explanation: **

**15**. If the numbers n – 2, 4n – 1 and 5n +2 are in AP, then the value of n is:

(A) 1

(B) 2

(C) − 1

(D) − 2

**Answer:**** (A)**

**Explanation:**

Let

a = n – 2

b = 4n – 1

c = 5n + 2

Since the terms are in A.P,

Therefore,

2b = a + c

⇒ 2 (4n – 1) = n – 2 + 5n + 2

⇒ 8n – 2 = 6n

⇒ 2n = 2

⇒ n = 1

**All the above questions can also be downloaded in PDF from the following link: **

**Also check: CBSE Class 10 Maths MCQs - All Chapters**

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