# MCQ Questions for Class 10 Maths Arithmetic Progressions with Answers

## 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
âˆ´ an+1 = 3 + 4(n + 1) = 7 + 4n
âˆ´ 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.
âˆ´ (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
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

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
âˆ´ 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
âˆ´ 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
âˆ´ a1 = 7, a2 - 10, a3 = 13
âˆ´ a= 7, d = 10 - 7 = 3
âˆ´ S12 = $\frac{12}{2}$ [2 Ã— 7 + (12 - 1) Ã—3] = 6[14 + 33] = 6 Ã— 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
âˆ´ Sum = $\frac{45}{2}$ [11 + 99] = $\frac{45}{2}$ Ã— 110 = 45 Ã— 55 = 2475

9. The sum of first n odd natural numbers is
(a) 2nÂ²
(b) 2n + 1
(c) 2n - 1
(d) nÂ²

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

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
âˆ´ ap + q = m
ap - 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´ an = $\frac{m+n}{2}$

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
âˆ´ 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.
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

Explaination:Reason: The numbers are 3, 9,15, 21, ..., 99
Here a = 3, d = 6 and an = 99
âˆ´ an = 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 + an] = $\frac{17}{2}$ [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

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 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
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´ 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:
âˆµ p, q, r are in AP.
âˆ´ 2q = p + r
â‡’ p + r - 2q = 0
âˆ´ p3 + r3 + (-2p)3 = 3 Ã— p Ã— r Ã— -2q
[Using ifa + 6 + c = 0 â‡’ a3 + b3 + c3 = 3 abc]
â‡’ 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]
(Ð°) 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
âˆµ an = 0
â‡’ a + (n - 1)d=0
â‡’ a + (5 - 1)(- 2) = 0
â‡’ 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 Ã— 3 = 15

27. The next term of the AP âˆš18, âˆš50, âˆš98, ........ is
(a) âˆš146
(b) âˆš128
(c) âˆš162
(d) âˆš200

Explaination:
(c) âˆš18, âˆš50, âˆš98, ..... = 3âˆš2, 5âˆš2, 7âˆš2, ......
âˆ´ Next term is 9âˆš2 = âˆš162

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 Ã— 1 + 1 = 3,
a2 = 2 Ã— 2 + 1 = 5,
a3 = 2 Ã— 3 + l= 7
âˆ´ 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
â‡’ S31 = $\frac{31}{2}$ Ã— 2(a + 15d)
â‡’ S31 = 31m

31. The first term of an AP of consecutive integers is pÂ² + 1. The sum of 2p + 1 terms of this AP is
(a) (p + 1)Â²
(b) (2p + 1) (p + 1)Â²
(c) (p+1)3
(d) p3 + (p + 1)3

Explaination:

32. If the sum of first n terms of an AP is An + BnÂ² 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 + BnÂ²
S1 = A Ã— 1 +B Ã— 1Â²
= A + B
âˆµ S1 = a1
âˆ´ a1 = A + B ... (1)
and S2 = A Ã— 2 + B Ã— 2Â²
â‡’ a1 + a2 = 2A + 4B
â‡’ (A + B) + a2 = 2A + 4B[Using (i)]
â‡’ a2 = A + 3B
âˆ´ d = a2 - a1 = 2B

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

Explaination:
âˆµ p - 1, p + 3 and 3p - 1 are in AP.
âˆ´ 2(p + 3) = p - 1 + 3p - 1
â‡’ 2p + 6 = 4/> -2.
â‡’ -2p = -8
â‡’ 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 â‰¥ 2

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

(ii) Tn = 3n+1
â‡’ T1 = 31+1 = 9,
T2 = 32+1 = 27,
T3= 33+1 = 81,
T4 = 34+1 = 243
âˆ´ Ist four terms are 9, 27, 81 and 243

(iii) T1 = 2, Tn = Tn-1 + 5, n â‰¥ 2
â‡’ 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
âˆ´ 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 Ã— 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
â‡’ a + (7 - 1)d = 4
[ âˆµ nth term of an AP = an = a + (n - 1)d]
â‡’ a + 6d = 4
â‡’ a + 6 Ã— (-4) = 4
â‡’ a - 24 = 4.
â‡’ a = 4 + 24 = 28
âˆ´ First term, a = 28

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

Explaination:
Here, a = 21, J= 18-21 = -3
Let an = 0
â‡’ a + (n - 1 )d = 0
â‡’ 21 +(n - 1)(-3) = 0
â‡’ (n - 1)(- 3) = -21
â‡’ n - 1 = $\frac{-21}{-3}$
â‡’ n = 8
âˆ´ 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
â‡’ 13 - 2p - 1 = 5p - 3 - 13
â‡’ 28 = 7p
â‡’ p = 4.

40. What is the common difference of an A.P. in which a21 - a7 = 84? [AI 2017]

Explaination:
Let â€˜cf be the common difference of the AP whose first term is â€˜aâ€™
Now, a21 - a7 = 84
â‡’ (a + 20d) - (a + 6d)= 84
â‡’ 20d - 6d = 84
â‡’ 14d = 84
â‡’ d = 6

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NCERT Solutions Very Good I Like It
• laxman
• 2020-February-17 15:31:16
good
• Bhanusam.manjula
• 2019-August-17 12:35:47
Ssc 2007batch regular halltichet number
• Ramu
• 2019-July-31 12:35:35
NCERT Solutions
• Kalamata gowriswari
• 2019-June-17 17:57:24
Open 10th result 2019ap
• Nagra rayal
• 2019-May-17 10:35:13
Yappudu 10th reselts
• Yanati nagarani
• 2019-May-17 10:20:31
10th result
• NAGESWARA
• 2019-May-11 18:30:50
How can see ssc ap 10th results
• m.kalyani
• 2019-May-11 18:30:36
10th result
• Naveen
• 2019-May-10 13:17:17
10th results
• Sri kesav
• 2019-May-10 09:58:50
Please tell me the correct date of AP ssc results
• K. Sudhakar
• 2019-May-10 09:58:43
Plz 10th result
• V pawan durga sriniva
• 2019-May-07 14:24:49
• V pawan durga sriniva
• 2019-May-07 14:24:39
• Charan
• 2019-May-07 14:24:21
Result ssc
• Ayesha
• 2019-May-07 14:24:17
At. What time did exam results will come?
• 2019-May-04 14:44:06
Results
• VANKUDOTH pavan kalyan
• 2018-May-30 10:16:07
Results
• C H VAISHAVI
• 2018-May-29 18:18:18
Cbse result by name
• varunsai sanne
• 2018-May-29 16:00:17
Xth results 2018
• varunsai sanne
• 2018-May-29 15:44:24
Xth results 2018
• Yalagala Srihari
• 2018-May-29 13:04:27
Exam results
• Alpesh Dhamale
• 2018-May-29 11:36:50
Nice article my baby wait for exam result i hope this year pass percentage may be increase
• m. mohan rao
• 2018-May-28 11:03:47
hi, s me cbse 10 class 2018 results
• akshaya
• 2018-May-26 10:22:07
at what time resilts are going to today
• Velupala Ashwarya
• 2018-May-15 11:01:52
My result
• Ramu
• 2018-May-12 16:09:34
good information