Free PDF Download of CBSE Class 10 Maths Chapter 5 Arithmetic Progressions Multiple Choice Questions with Answers. MCQ Questions for Class 10 MathsÂ with Answers was Prepared Based on Latest Exam Pattern. Students can solve NCERT Class 10 Maths Arithmetic ProgressionsMCQs with Answers to know their preparation level.

## Class 10 Maths MCQs Chapter 5 Arithmetic Progressions

**Arithmetic Progression Class 10 MCQ**

1. The n^{th} term of an A.P. is given by a_{n} = 3 + 4n. The common difference is

(a) 7

(b) 3

(c) 4

(d) 1

**Answer/Explanation**

Answer: c

Explaination:Reason: We have an = 3 + 4n

âˆ´ a_{n+1} = 3 + 4(n + 1) = 7 + 4n

âˆ´ d = a_{n+1} – a_{n}

= (7 + 4n) – (3 + 4n)

= 7 – 3

= 4

**Arithmetic Progression MCQ Class 10**

2. If p, q, r and s are in A.P. then r – q is

(a) s – p

(b) s – q

(c) s – r

(d) none of these

**Answer/Explanation**

Answer: c

Explaination:Reason: Since p, q, r, s are in A.P.

âˆ´ (q – p) = (r – q) = (s – r) = d (common difference)

**Ap MCQ Class 10**

3. If the sum of three numbers in an A.P. is 9 and their product is 24, then numbers are

(a) 2, 4, 6

(b) 1, 5, 3

(c) 2, 8, 4

(d) 2, 3, 4

**Answer/Explanation**

Answer: d

Explaination:Reason: Let three numbers be a – d, a, a + d

âˆ´ a – d +a + a + d = 9

â‡’ 3a = 9

â‡’ a = 3

Also (a – d) . a . (a + d) = 24

â‡’ (3 -d) .3(3 + d) = 24

â‡’ 9 – dÂ² = 8

â‡’ dÂ² = 9 – 8 = 1

âˆ´ d = Â± 1

Hence numbers are 2, 3, 4 or 4, 3, 2

**MCQ On Arithmetic Progression Class 10**

4. The (n – 1)^{th} term of an A.P. is given by 7,12,17, 22,… is

(a) 5n + 2

(b) 5n + 3

(c) 5n – 5

(d) 5n – 3

**Answer/Explanation**

Answer: d

Explaination:Reason: Here a = 7, d = 12-7 = 5

âˆ´ a_{n-1} = a + [(n – 1) – l]d = 7 + [(n – 1) -1] (5) = 7 + (n – 2)5 = 7 + 5n – 10 = 5M – 3

**Class 10 Arithmetic Progression MCQ**

5. The n^{th} term of an A.P. 5, 2, -1, -4, -7 … is

(a) 2n + 5

(b) 2n – 5

(c) 8 – 3n

(d) 3n – 8

**Answer/Explanation**

Answer: c

Explaination:Reason: Here a = 5, d = 2 – 5 = -3

a_{n} = a + (n – 1)d = 5 + (n – 1) (-3) = 5 – 3n + 3 = 8 – 3n

**Arithmetic Progression MCQ Class 10**

6. The 10^{th} term from the end of the A.P. -5, -10, -15,…, -1000 is

(a) -955

(b) -945

(c) -950

(d) -965

**Answer/Explanation**

Answer: a

Explaination:Reason: Here l = -1000, d = -10 – (-5) = -10 + 5 = – 5

âˆ´ 10^{th} term from the end = l – (n – 1 )d = -1000 – (10 – 1) (-5) = -1000 + 45 = -955

**Ap Class 10 MCQ**

7. Find the sum of 12 terms of an A.P. whose nth term is given by a_{n} = 3n + 4

(a) 262

(b) 272

(c) 282

(d) 292

**Answer/Explanation**

Answer: a

Explaination:Reason: Here a_{n} = 3n + 4

âˆ´ a_{1} = 7, a_{2} – 10, a_{3} = 13

âˆ´ a= 7, d = 10 – 7 = 3

âˆ´ S_{12} = \(\frac{12}{2}\)[2 Ã— 7 + (12 – 1) Ã—3] = 6[14 + 33] = 6 Ã— 47 = 282

**Arithmetic Progression Class 10 MCQ With Answers**

8. The sum of all two digit odd numbers is

(a) 2575

(b) 2475

(c) 2524

(d) 2425

**Answer/Explanation**

Answer: b

Explaination:Reason: All two digit odd numbers are 11,13,15,… 99, which are in A.P.

Since there are 90 two digit numbers of which 45 numbers are odd and 45 numbers are even

âˆ´ Sum = \(\frac{45}{2}\)[11 + 99] = \(\frac{45}{2}\) Ã— 110 = 45 Ã— 55 = 2475

**Class 10 Ap MCQ**

9. The sum of first n odd natural numbers is

(a) 2nÂ²

(b) 2n + 1

(c) 2n – 1

(d) nÂ²

**Answer/Explanation**

Answer: d

Explaination:Reason: Required Sum = 1 + 3 + 5 + … + upto n terms.

Here a = 1, d = 3 – 1 = 2

Sum = \(\frac{n}{2}\)[2 Ã— 1 + (n – 1) Ã— 2] = \(\frac{n}{2}\)[2 + 2n – 2] = \(\frac{n}{2}\) Ã— 2n = nÂ²Reason: All two digit odd numbers are 11,13,15,… 99, which are in A.P.

Since there are 90 two digit numbers of which 45 numbers are odd and 45 numbers are even

âˆ´ Sum = \(\frac{45}{2}\)[11 + 99] = \(\frac{45}{2}\) Ã— 110 = 45 Ã— 55 = 2475

**Arithmetic Progression MCQ Questions Class 10**

10. If (p + q)^{th} term of an A.P. is m and (p – q)^{tn} term is n, then pth term is

**Answer/Explanation**

Answer: d

Explaination:Reason: Let a is first term and d is common difference

âˆ´ a_{p + q} = m

a_{p – q} = n

â‡’ a + (p + q – 1)d = m = …(i)

â‡’ a + (p – q – 1)d = m = …(ii)

On adding (i) and (if), we get

2a + (2p – 2)d = m + n

â‡’ a + (p -1)d = \(\frac{m+n}{2}\) …[Dividing by 2

âˆ´ a_{n} = \(\frac{m+n}{2}\)

**MCQ Of Arithmetic Progression For Class 10**

11. If a, b, c are in A.P. then \(\frac{a-b}{b-c}\) is equal to

**Answer/Explanation**

Answer: a

Explaination:Reason: Since a, b, c are in A.P.

âˆ´ b – a = c – b

â‡’ \(\frac{b-a}{c-b}\) = 1

â‡’ \(\frac{a-b}{b-c}\) = 1

**Arithmetic Progression MCQ Questions And Answers**

12. The number of multiples lie between n and nÂ² which are divisible by n is

(a) n + 1

(b) n

(c) n – 1

(d) n – 2

**Answer/Explanation**

Answer: d

Explaination:Reason: Multiples of n from 1 to nÂ²Â are n Ã— 1, n Ã— 2, n Ã— 3, …, mÃ— n

âˆ´ There are n numbers

Thus, the number of mutiples of n which lie between n and nÂ² is (n – 2) leaving first and last in the given list: Total numbers are (n – 2).

**Arithmetic Progression Class 10 MCQs**

13. If a, b, c, d, e are in A.P., then the value of a – 4b + 6c – 4d + e is

(a) 0

(b) 1

(c) -1.

(d) 2

**Answer/Explanation**

Answer: a

Explaination:Reason: Let common difference of A.P. be x

âˆ´ b = a + x, c = a + 2x, d = a + 3x and e = a + 4x

Given equation n-4b + 6c-4d + c

= a – 4(a + x) + 6(A + 2r) – 4(n + 3x) + (o + 4.v)

= a – 4a – 4x + 6a + 12x – 4a – 12x + a + 4x = 8a – 8a + 16x – 16x = 0

**MCQ Questions For Class 10 Maths Arithmetic Progression**

14. The next term of the sequence

**Answer/Explanation**

Answer: a

Explaination:

**MCQ Questions On Arithmetic Progression For Class 10**

15. n^{th} term of the sequence a, a + d, a + 2d,… is

(a) a + nd

(b) a – (n – 1)d

(c) a + (n – 1)d

(d) n + nd

**Answer/Explanation**

Answer: a

Explaination:Reason: an = a + (n – 1)d

16. The 10th term from the end of the A.P. 4, 9,14, …, 254 is

(a) 209

(b) 205

(c) 214

(d) 213

**Answer/Explanation**

Answer: a

Explaination:Reason: Here l – 254, d = 9-4 = 5

âˆ´ 10^{th} term from the end = l – (10 – 1 )d = 254 -9d = 254 = 9(5) = 254 – 45 = 209

17. If 2x, x + 10, 3x + 2 are in A.P., then x is equal to

(a) 0

(b) 2

(c) 4

(d) 6

**Answer/Explanation**

Answer: d

Explaination:Reason: Since 2x, x + 10 and 3x + 2 are in A.P.

âˆ´ 2(x + 10) = 2x + (3x + 2)

â‡’ 2x + 20 – 5x + 2

â‡’ 2x – 5x = 2 – 20

â‡’ 3x = 18

â‡’ x = 6

18. The sum of all odd integers between 2 and 100 divisible by 3 is

(a) 17

(b) 867

(c) 876

(d) 786

**Answer/Explanation**

Answer: b

Explaination:Reason: The numbers are 3, 9,15, 21, …, 99

Here a = 3, d = 6 and a_{n} = 99

âˆ´ a_{n} = a + (n – 1 )d

â‡’ 99 = 3 + (n – 1) x 6

â‡’ 99 = 3 + 6n – 6

â‡’ 6n = 102

â‡’ n = 17

Required Sum = \(\frac{n}{2}\)[a + a_{n}] = \(\frac{17}{2}\)[3 + 99] = \(\frac{17}{2}\) Ã— 102 = 867

19. If the numbers a, b, c, d, e form an A.P., then the value of a – 4b + 6c – 4d + e is

(a) 0

(b) 1

(c) -1

(d) 2

**Answer/Explanation**

Answer: a

Explaination:Reason: Let x be the common difference of the given AP

âˆ´ b = a + x, c = a + 2x, d = a + 3x and e = a + 4x

âˆ´ a – 4b + 6c – 4d + e = a – 4 (a + x) + 6(a + 2x) – 4(a + 3x) + (a + 4x)

= a – 4a – 4x + 6a + 12x – 4a – 12x + a + 4x = 8a – 8a + 16x – 16x = 0

20. If 7 times the 7^{th} term of an A.P. is equal to 11 times its 11^{th} term, then 18^{th} term is

(a) 18

(b) 9

(c) 77

(d) 0

**Answer/Explanation**

Answer: d

Explaination:Reason: We have 7a_{7} = 11a_{11}

â‡’ 7[a + (7 – 1)d] = 11[a + (11 – 1 )d]

â‡’ 7(a + 6d) = 11(a + 10d)

â‡’ 7a + 42d = 11a + 110d

â‡’ 4a = -68d

â‡’ a = -17d

âˆ´ a_{18} = a + (18 – 1)d = a + 17d = -17d + 17d = 0

We hope the given MCQ Questions for Class 10 Maths Arithmetic Progressions with Answers will help you. If you have any query regarding CBSE Class 10 Maths Chapter 5 Arithmetic Progressions Multiple Choice Questions with Answers, drop a comment below and we will get back to you at the earliest.