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 for the CBSE Term-I Board Exam that will include only the objective type questions. Answers to all questions have been provided with detailed explanations.

**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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