MCQ Questions for Class 10 Maths Arithmetic Progressions with Answers

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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

Answer/Explanation

Answer: c
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

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
∴ 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

Answer/Explanation

Answer: c
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

Answer/Explanation

Answer: a
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

Answer/Explanation

Answer: a
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

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
MCQ Questions for Class 10 Maths Arithmetic Progressions with Answers 1

Answer/Explanation

Answer: d
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
MCQ Questions for Class 10 Maths Arithmetic Progressions with Answers 2

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
MCQ Questions for Class 10 Maths Arithmetic Progressions with Answers 3

Answer/Explanation

Answer: a
Explaination:
MCQ Questions for Class 10 Maths Arithmetic Progressions with Answers 4


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

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
∴ 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

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 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

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 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

Answer/Explanation

Answer: d
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


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