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## Class 10 Maths MCQs Chapter 5 Arithmetic Progressions

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

∴ a_{n+1} = 3 + 4(n + 1) = 7 + 4n

∴ d = a_{n+1} – a_{n}

= (7 + 4n) – (3 + 4n)

= 7 – 3

= 4

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.

∴ (q – p) = (r – q) = (s – r) = d (common difference)

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

∴ a – d +a + a + d = 9

⇒ 3a = 9

⇒ a = 3

Also (a – d) . a . (a + d) = 24

⇒ (3 -d) .3(3 + d) = 24

⇒ 9 – d² = 8

⇒ d² = 9 – 8 = 1

∴ d = ± 1

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

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

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

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

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

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

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

∴ a_{1} = 7, a_{2} – 10, a_{3} = 13

∴ a= 7, d = 10 – 7 = 3

∴ S_{12} = \(\frac{12}{2}\)[2 × 7 + (12 – 1) ×3] = 6[14 + 33] = 6 × 47 = 282

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

∴ Sum = \(\frac{45}{2}\)[11 + 99] = \(\frac{45}{2}\) × 110 = 45 × 55 = 2475

9. The sum of first n odd natural numbers is

(a) 2n²

(b) 2n + 1

(c) 2n – 1

(d) n²

**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 × 1 + (n – 1) × 2] = \(\frac{n}{2}\)[2 + 2n – 2] = \(\frac{n}{2}\) × 2n = n²Reason: 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

∴ Sum = \(\frac{45}{2}\)[11 + 99] = \(\frac{45}{2}\) × 110 = 45 × 55 = 2475

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

∴ a_{p + q} = m

a_{p – q} = n

⇒ a + (p + q – 1)d = m = …(i)

⇒ a + (p – q – 1)d = m = …(ii)

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

2a + (2p – 2)d = m + n

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

∴ a_{n} = \(\frac{m+n}{2}\)

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.

∴ b – a = c – b

⇒ \(\frac{b-a}{c-b}\) = 1

⇒ \(\frac{a-b}{b-c}\) = 1

12. The number of multiples lie between n and n² 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 n² are n × 1, n × 2, n × 3, …, m× n

∴ There are n numbers

Thus, the number of mutiples of n which lie between n and n² is (n – 2) leaving first and last in the given list: Total numbers are (n – 2).

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

∴ 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

14. The next term of the sequence

**Answer/Explanation**

Answer: a

Explaination:

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

∴ 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.

∴ 2(x + 10) = 2x + (3x + 2)

⇒ 2x + 20 – 5x + 2

⇒ 2x – 5x = 2 – 20

⇒ 3x = 18

⇒ 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

∴ a_{n} = a + (n – 1 )d

⇒ 99 = 3 + (n – 1) x 6

⇒ 99 = 3 + 6n – 6

⇒ 6n = 102

⇒ n = 17

Required Sum = \(\frac{n}{2}\)[a + a_{n}] = \(\frac{17}{2}\)[3 + 99] = \(\frac{17}{2}\) × 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

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

∴ 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}

⇒ 7[a + (7 – 1)d] = 11[a + (11 – 1 )d]

⇒ 7(a + 6d) = 11(a + 10d)

⇒ 7a + 42d = 11a + 110d

⇒ 4a = -68d

⇒ a = -17d

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

21. If p, q, r are in AP, then p^{3} + r^{3} – 8q^{3} is equal to

(a) 4pqr

(b) -6pqr

(c) 2pqr

(d) 8pqr

**Answer/Explanation**

Answer: b

Explaination:

∵ p, q, r are in AP.

∴ 2q = p + r

⇒ p + r – 2q = 0

∴ p^{3} + r^{3} + (-2p)^{3} = 3 × p × r × -2q

[Using ifa + 6 + c = 0 ⇒ a^{3} + b^{3} + c^{3} = 3 abc]

⇒ p^{3} + r^{3} – 8q^{3} = -6pqr.

22. In an AP, if a = 3.5, d = 0, n = 101, then a will be [NCERT Exemplar Problems]

(a) 0

(b) 3.5

(c) 103.5

(d) 104.5

**Answer/Explanation**

Answer: b

Explaination: (b) a_{101} = 3.5 + 0(100) = 3.5

23. The list of numbers -10, -6, -2, 2, … is [NCERT Exemplar Problems]

(а) an AP with d = -16

(b) an AP with d = 4

(c) an AP with d = -4

(d) not an AP

**Answer/Explanation**

Answer: b

Explaination: (b) An AP with d = 4.

24. 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 [NCERT Exemplar Problems]

(a) -1

(b) -8

(c) 7

(d) -9

**Answer/Explanation**

Answer: c

Explaination:

a_{4} – b_{4} = (a_{1} + 3d) – (b_{1} + 3d)

= a_{1} – b_{1}= – 1 – (-8) = 7

25. In an AP, if d = -2, n = 5 and an = 0, the value of a is

(a) 10

(b) 5

(c) -8

(d) 8

**Answer/Explanation**

Answer: d

Explaination:

d = – 2, n = 5, a_{n} = 0

∵ a_{n} = 0

⇒ a + (n – 1)d=0

⇒ a + (5 – 1)(- 2) = 0

⇒ a = 8

Correct option is (d).

26. If the common difference of an AP is 3, then a_{20} – a_{15} is

(a) 5

(b) 3

(c) 15

(d) 20

**Answer/Explanation**

Answer: c

Explaination:

Common difference, d = 3

a_{20} – a_{15} = (a + 19d) – (a+ 14d)

= 5d=5 × 3 = 15

27. The next term of the AP √18, √50, √98, …….. is

(a) √146

(b) √128

(c) √162

(d) √200

**Answer/Explanation**

Answer: c

Explaination:

(c) √18, √50, √98, ….. = 3√2, 5√2, 7√2, ……

∴ Next term is 9√2 = √162

28. The common difference of the AP

(a) p

(b) -p

(c) -1

(d) 1

**Answer/Explanation**

Answer: c

Explaination: (c) Common difference = a_{2} – a_{1}

29. If the n^{th} term of an AP is (2n +1), then the sum of its first three terms is

(a) 6n + 3

(b) 15

(c) 12

(d) 21

**Answer/Explanation**

Answer: b

Explaination:

a_{1}= 2 × 1 + 1 = 3,

a_{2} = 2 × 2 + 1 = 5,

a_{3} = 2 × 3 + l= 7

∴ Sum = 3 + 5 + 7 = 15

30. An AP consists of 31 terms. If its 16th term is m, then sum of all the terms of this AP is

(a) 16 m

(b) 47 m

(c) 31 m

(d) 52 m

**Answer/Explanation**

Answer: c

Explaination:

S_{31} =\(\frac{31}{2}\) (2a + 30d)

a_{16} = a + 15d = zw

⇒ S_{31} = \(\frac{31}{2}\) × 2(a + 15d)

⇒ S_{31} = 31m

31. The first term of an AP of consecutive integers is p² + 1. The sum of 2p + 1 terms of this AP is

(a) (p + 1)²

(b) (2p + 1) (p + 1)²

(c) (p+1)^{3}

(d) p^{3} + (p + 1)^{3}

**Answer/Explanation**

Answer: d

Explaination:

32. If the sum of first n terms of an AP is An + Bn² where A and B are constants, the common difference of AP will be

(a) A + B

(b) A – B

(c) 2A

(d) 2B

**Answer/Explanation**

Answer: d

Explaination:

S_{n} = An + Bn²

S_{1} = A × 1 +B × 1²

= A + B

∵ S_{1} = a_{1}

∴ a_{1} = A + B … (1)

and S_{2} = A × 2 + B × 2²

⇒ a_{1} + a_{2} = 2A + 4B

⇒ (A + B) + a_{2} = 2A + 4B[Using (i)]

⇒ a_{2} = A + 3B

∴ d = a_{2} – a_{1} = 2B

33. If p – 1, p + 3, 3p – 1 are in AP, then p is equal to ______ .

**Answer/Explanation**

Answer:

Explaination:

∵ p – 1, p + 3 and 3p – 1 are in AP.

∴ 2(p + 3) = p – 1 + 3p – 1

⇒ 2p + 6 = 4/> -2.

⇒ -2p = -8

⇒ p = 4.

34. Write down the first four terms of the sequences whose general terms are

(i) T_{n} = 2n + 3

(ii) T_{n} =3^{n + 1}

(iii) T_{1} = 2, T_{n} = T_{n-1}+ 5, n ≥ 2

**Answer/Explanation**

Answer:

Explaination:

(i) T_{n}= 2n + 3

T_{1} = 2 × 1 + 3 = 5,

T_{2} = 2 × 2 + 3 = 7,

T_{3} = 2 × 3 + 3 = 9,

T_{4} = 2 × 4 + 3 = 11

∴ Ist four terms are 5, 7, 9 and 11.

(ii) T_{n} = 3^{n+1}

⇒ T_{1} = 3^{1+1} = 9,

T_{2} = 3^{2+1} = 27,

T_{3}= 3^{3+1} = 81,

T_{4} = 3^{4+1} = 243

∴ Ist four terms are 9, 27, 81 and 243

(iii) T_{1} = 2, T_{n} = T_{n-1} + 5, n ≥ 2

⇒ T_{2} = T_{2-1} + 5

= T_{1} + 5 = 2 + 5 = 7

T_{3} = T_{3-1} + 5

= T_{2} + 5 = 7 + 5 = 12

and T_{4}=T_{4-1} + 5

= T_{3} + 5 = 12 + 5 = 17

∴ Ist four terms are 2, 7, 12 and 17.

35. Find:

The 10th term of 10.0, 10.5, 11.0, 11.5, …..

**Answer/Explanation**

Answer:

Explaination:

a=10, d = 10.5 – 10 = 0.5

a_{10} = a + 9d= 10 + 9 × 0.5 = 14.5

36. In an A.P., if the common difference (d) = – 4 and the seventh term (a_{7}) is 4, then find the first term. [CBSE 2018]

**Answer/Explanation**

Answer:

Explaination:

Let a = first term,

Given, d = – 4, a_{7} = 4

⇒ a + (7 – 1)d = 4

[ ∵ nth term of an AP = a_{n} = a + (n – 1)d]

⇒ a + 6d = 4

⇒ a + 6 × (-4) = 4

⇒ a – 24 = 4.

⇒ a = 4 + 24 = 28

∴ First term, a = 28

37. Write the n^{th} term of the A.P. [Delhi 2017 (C)]

**Answer/Explanation**

Answer:

Explaination:

38. Which term of the AP 21, 18, 15, … , is zero?

**Answer/Explanation**

Answer:

Explaination:

Here, a = 21, J= 18-21 = -3

Let a_{n} = 0

⇒ a + (n – 1 )d = 0

⇒ 21 +(n – 1)(-3) = 0

⇒ (n – 1)(- 3) = -21

⇒ n – 1 = \(\frac{-21}{-3}\)

⇒ n = 8

∴ 8th term is zero.

39. For what value ofp, are 2p+ 1, 13, 5p – 3 three consecutive terms of an AP?

**Answer/Explanation**

Answer:

Explaination:

If terms are in AP, then

13 – (2p + 1) = (5p – 3) – 13

⇒ 13 – 2p – 1 = 5p – 3 – 13

⇒ 28 = 7p

⇒ p = 4.

40. What is the common difference of an A.P. in which a_{21} – a_{7} = 84? [AI 2017]

**Answer/Explanation**

Answer:

Explaination:

Let ‘cf be the common difference of the AP whose first term is ‘a’

Now, a_{21} – a_{7} = 84

⇒ (a + 20d) – (a + 6d)= 84

⇒ 20d – 6d = 84

⇒ 14d = 84

⇒ d = 6

41. The first term of an AP is p and its common difference is q. Find its 10th term.

**Answer/Explanation**

Answer:

Explaination:

∵ a = p and d = q

a_{n} = a + (n – 1)d

⇒ a_{10} =p + 9q

42. Which term of the AP 14, 11, 8, ….is -1?

**Answer/Explanation**

Answer:

Explaination:

Here, a = 14, d= 11 – 14 = – 3

Let a_{n} = – 1

⇒ a + (n – 1 )d = – 1

⇒ 14 + (n – 1)(- 3) = – 1

⇒ (n – 1)(- 3) = – 1 – 14

⇒ n – 1 = \(\frac{-15}{-3}\) = 5

⇒ n = 6

∴ 6th term of the AP is -1.

43. Write the next two terms of the AP: 1,-1, -3, -5, …

**Answer/Explanation**

Answer:

Explaination:

a_{1} = 1

a_{1} = a_{2} – a_{1} = – 1 – 1 = – 2

a_{5} = a_{1} + 4d

= 1 + 4(-2)

= 1 – 8 = — 7

a_{6} = a_{5} + d = – 7 – 2 = – 9

Next two terms are – 7 and – 9.

44. If a_{n} = \(\frac{n(n-3)}{n+4}\), then find 18th term of this sequence.

**Answer/Explanation**

Answer:

Explaination:

45. If the first term of an AP is 2 and common difference is 4, then sum of its first 40 terms is ______ .

**Answer/Explanation**

Answer:

Explaination:

S_{40}(2 × 2 + 39 × 4)

= 40 × (80)= 3200

46. Three numbers in an AP have sum 24. Its middle term is ____ .

**Answer/Explanation**

Answer:

Explaination:

Let numbers be a – d, a and a + d.

⇒ a – d+a + a + d= 24

⇒ 3a = 24

⇒ a = 8

∴ Middle term = 8.

47. The value of the expression 1 – 6 + 2 – 7 + 3 – 8 + ….. to 100 terms is ______ .

**Answer/Explanation**

Answer:

Explaination:

1 – 6 + 2 – 7 + 3 – 8 + … 100 terms

= (1 + 2 + 3 +… up to 50 terms) + (-6 -7 – 8 – … up to 50 terms)

= \(\frac{50}{2}\)[2 × 1 + 49 × 1] + [\(\frac{50}{2}\){2 × (-6) + 49 × (-1)}]

= 25 × 51 + 25 x (-12 – 49) = 25(51 – 61)

= -250.

48. If the sum of first m terms of an AP is 2m² + 3m, then what is its second term?

**Answer/Explanation**

Answer:

Explaination:

S_{m = 2m² + 3m
∴ S1 = 2 × 1² + 3 × 1 = 5 = a1
and S2 = 2 × 2² + 3 × 2 = 14 …(i)
⇒ a1 + a2 = 14
⇒ 5 + a2 = 14
⇒ a2 = 9}

49. If the sum of first p terms of an AP is ap² + bp, find its common difference.

**Answer/Explanation**

Answer:

Explaination:

S_{p = ap² + bp
S1= a × 1² + b × 1
= a + b = a1
and S2 = a × 2² + b × 2 = 4a + 2b …(i)
⇒ a1 + a2 = 4a + 2b
⇒ a + b + a2 = 4a + 2b
⇒ a2 = 3a + b [Using eq. (i)]
Now, d = a2 – a1
= (3a + b)-(a + b) = 2a.}

50. If sum of first n terms of an AP is 2n² + 5n. Then find S_{20}.

**Answer/Explanation**

Answer:

Explaination:

S_{n = 2n² + 5n
S20 = 2(20)² + 5 × 20
= 2 × 400 + 100 = 900}

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