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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
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
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. 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
21. If p, q, r are in AP, then p3 + r3 – 8q3 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
∴ 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
Answer/Explanation
Answer: b
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
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:
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
Answer/Explanation
Answer: d
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
Answer/Explanation
Answer: c
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
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 = 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
Answer/Explanation
Answer: b
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
Answer/Explanation
Answer: c
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
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:
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 ______ .
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) Tn = 2n + 3
(ii) Tn =3n + 1
(iii) T1 = 2, Tn = Tn-1+ 5, n ≥ 2
Answer/Explanation
Answer:
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, …..
Answer/Explanation
Answer:
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]
Answer/Explanation
Answer:
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)]
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 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?
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 a21 – a7 = 84? [AI 2017]
Answer/Explanation
Answer:
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.
Answer/Explanation
Answer:
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?
Answer/Explanation
Answer:
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, …
Answer/Explanation
Answer:
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.
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:
S40(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:
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.
Answer/Explanation
Answer:
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.
Answer/Explanation
Answer:
Explaination:
Sn = 2n² + 5n
S20 = 2(20)² + 5 × 20
= 2 × 400 + 100 = 900
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