# MCQ Questions for Class 10 Maths Arithmetic Progressions with Answers

Free PDF Download of CBSE Class 10 Maths Chapter 5 Arithmetic Progressions Multiple Choice Questions with Answers. MCQ Questions for Class 10 Maths 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

1. The nth term of an A.P. is given by an = 3 + 4n. The common difference is
(a) 7
(b) 3
(c) 4
(d) 1

Explaination:Reason: We have an = 3 + 4n
∴ an+1 = 3 + 4(n + 1) = 7 + 4n
∴ d = an+1 – an
= (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

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

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

Explaination:Reason: Here a = 7, d = 12-7 = 5
∴ an-1 = a + [(n – 1) – l]d = 7 + [(n – 1) -1] (5) = 7 + (n – 2)5 = 7 + 5n – 10 = 5M – 3

5. The nth term of an A.P. 5, 2, -1, -4, -7 … is
(a) 2n + 5
(b) 2n – 5
(c) 8 – 3n
(d) 3n – 8

Explaination:Reason: Here a = 5, d = 2 – 5 = -3
an = a + (n – 1)d = 5 + (n – 1) (-3) = 5 – 3n + 3 = 8 – 3n

6. The 10th term from the end of the A.P. -5, -10, -15,…, -1000 is
(a) -955
(b) -945
(c) -950
(d) -965

Explaination:Reason: Here l = -1000, d = -10 – (-5) = -10 + 5 = – 5
∴ 10th 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 an = 3n + 4
(a) 262
(b) 272
(c) 282
(d) 292

Explaination:Reason: Here an = 3n + 4
∴ a1 = 7, a2 – 10, a3 = 13
∴ a= 7, d = 10 – 7 = 3
∴ S12 = $$\frac{12}{2}$$[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

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

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 Explaination:Reason: Let a is first term and d is common difference
∴ ap + q = m
ap – 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
∴ an = $$\frac{m+n}{2}$$

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

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

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 Explaination: 15. nth 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

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

Explaination:Reason: Here l – 254, d = 9-4 = 5
∴ 10th 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

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

Explaination:Reason: The numbers are 3, 9,15, 21, …, 99
Here a = 3, d = 6 and an = 99
∴ an = a + (n – 1 )d
⇒ 99 = 3 + (n – 1) x 6
⇒ 99 = 3 + 6n – 6
⇒ 6n = 102
⇒ n = 17
Required Sum = $$\frac{n}{2}$$[a + an] = $$\frac{17}{2}$$[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

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 7th term of an A.P. is equal to 11 times its 11th term, then 18th term is
(a) 18
(b) 9
(c) 77
(d) 0

Explaination:Reason: We have 7a7 = 11a11
⇒ 7[a + (7 – 1)d] = 11[a + (11 – 1 )d]
⇒ 7(a + 6d) = 11(a + 10d)
⇒ 7a + 42d = 11a + 110d
⇒ 4a = -68d
⇒ a = -17d
∴ a18 = a + (18 – 1)d = a + 17d = -17d + 17d = 0

21. If p, q, r are in AP, then p3 + r3 – 8q3 is equal to
(a) 4pqr
(b) -6pqr
(c) 2pqr
(d) 8pqr

Explaination:
∵ p, q, r are in AP.
∴ 2q = p + r
⇒ p + r – 2q = 0
∴ p3 + r3 + (-2p)3 = 3 × p × r × -2q
[Using ifa + 6 + c = 0 ⇒ a3 + b3 + c3 = 3 abc]
⇒ p3 + r3 – 8q3 = -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

Explaination: (b) a101 = 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

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

Explaination:
a4 – b4 = (a1 + 3d) – (b1 + 3d)
= a1 – b1= – 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

Explaination:
d = – 2, n = 5, an = 0
∵ an = 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 a20 – a15 is
(a) 5
(b) 3
(c) 15
(d) 20

Explaination:
Common difference, d = 3
a20 – a15 = (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

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

Explaination: (c) Common difference = a2 – a1

29. If the nth 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

Explaination:
a1= 2 × 1 + 1 = 3,
a2 = 2 × 2 + 1 = 5,
a3 = 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

Explaination:
S31 =$$\frac{31}{2}$$ (2a + 30d)
a16 = a + 15d = zw
⇒ S31 = $$\frac{31}{2}$$ × 2(a + 15d)
⇒ S31 = 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) p3 + (p + 1)3

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

Explaination:
Sn = An + Bn²
S1 = A × 1 +B × 1²
= A + B
∵ S1 = a1
∴ a1 = A + B … (1)
and S2 = A × 2 + B × 2²
⇒ a1 + a2 = 2A + 4B
⇒ (A + B) + a2 = 2A + 4B[Using (i)]
⇒ a2 = A + 3B
∴ d = a2 – a1 = 2B

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

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) Tn = 2n + 3
(ii) Tn =3n + 1
(iii) T1 = 2, Tn = Tn-1+ 5, n ≥ 2

Explaination:
(i) Tn= 2n + 3
T1 = 2 × 1 + 3 = 5,
T2 = 2 × 2 + 3 = 7,
T3 = 2 × 3 + 3 = 9,
T4 = 2 × 4 + 3 = 11
∴ Ist four terms are 5, 7, 9 and 11.

(ii) Tn = 3n+1
⇒ T1 = 31+1 = 9,
T2 = 32+1 = 27,
T3= 33+1 = 81,
T4 = 34+1 = 243
∴ Ist four terms are 9, 27, 81 and 243

(iii) T1 = 2, Tn = Tn-1 + 5, n ≥ 2
⇒ T2 = T2-1 + 5
= T1 + 5 = 2 + 5 = 7
T3 = T3-1 + 5
= T2 + 5 = 7 + 5 = 12
and T4=T4-1 + 5
= T3 + 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, …..

Explaination:
a=10, d = 10.5 – 10 = 0.5
a10 = a + 9d= 10 + 9 × 0.5 = 14.5

36. In an A.P., if the common difference (d) = – 4 and the seventh term (a7) is 4, then find the first term. [CBSE 2018]

Explaination:
Let a = first term,
Given, d = – 4, a7 = 4
⇒ a + (7 – 1)d = 4
[ ∵ nth term of an AP = an = 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 nth term of the A.P. [Delhi 2017 (C)] Explaination: 38. Which term of the AP 21, 18, 15, … , is zero?

Explaination:
Here, a = 21, J= 18-21 = -3
Let an = 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?

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 a21 – a7 = 84? [AI 2017]

Explaination:
Let ‘cf be the common difference of the AP whose first term is ‘a’
Now, a21 – a7 = 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.

Explaination:
∵ a = p and d = q
an = a + (n – 1)d
⇒ a10 =p + 9q

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

Explaination:
Here, a = 14, d= 11 – 14 = – 3
Let an = – 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, …

Explaination:
a1 = 1
a1 = a2 – a1 = – 1 – 1 = – 2
a5 = a1 + 4d
= 1 + 4(-2)
= 1 – 8 = — 7
a6 = a5 + d = – 7 – 2 = – 9
Next two terms are – 7 and – 9.

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

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

Explaination:
S40(2 × 2 + 39 × 4)
= 40 × (80)= 3200

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

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 ______ .

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?

Explaination:
Sm = 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.

Explaination:
Sp = 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 S20.