Arithmetic Progressions – Class 10

CBSE CLASS 10 MATHEMATICS
Arithmetic Progressions – Chapter 5 /MCQ

1. The 10 th term of the AP: 2, 7, 12……………. is —————-
   A. 43
   B. 45
   C. 47
2. For what values of k will k+3, 2k +1, 2k +9 is the consecutive terms of an AP?
   A. 8
   B. 10
   C. 12
3. Which term of the AP 1125, 1100, 1075, 1050 ……….. is zero?
   A. 45
   b. 46
   c. 47
4. In the following AP, find the missing term: 2, ———, 26
   A. 12
   B. 13
   C. 14
5. Which term of the AP: 4, 7, 10, 13 ………….. is 76?
   A. 20
   B. 25
   C. 30
6. The sum of first 500 positive integers is given by ———–
   A. 125125
   B. 125250
   C. 125275
7. If the sum of three numbers in AP is 72, then the middle term is ———–
   A. 20
   B. 24
   C. 28
8. If the n th term of an AP is (3n +2), then the sum of its first three terms is —————-
   A. 11
   B. 24
   C. 35
9. The 11 th term from the last term (towards the first term) of the
   AP: 10, 7, 4, ——— -62 is —————-
   A. 25
   B. 32
   C. -32
10. Which term of the AP: 84, 77, 70, 63 ———– is zero?
    A. 11
    B. 12
    C. 13

Answers with explanation:

1. 47
   (Explanation: Here a = 2, d=5, n=10
   n th term = a + (n-1)d = 2 + (10 – 1) 5 = 2 + 45 = 47)
2. k = 10
   (Explanation: 2k + 1 = ½ (k + 3 + 2k + 9)
   2k + 1 = ½ (3k + 12)
   4k + 2 = 3k + 12
   4k – 3k = 12 -2
   K = 10)
3. 46
   (Explanation: Given n th term = 0
   i.e. a + (n -1) d = 1125 + (n – 1) (-25) =0
   1125 – 25n +25 = 0
   1150 – 25 n = 0
   1150 = 25n
   n = 1150/25 = 46.)
4. 14
   (Explanation: Middle term = ½ (26 +2) = 14.)
5. 25
   (Explanation: Given n th term = 76
   4 + (n – 1) 3 = 76
   4 + 3n – 3 = 76
   1 + 3n = 76
   3n = 76 – 1 = 75
   n = 75/3 = 25)
6. 125250.
   (Explanation: The sum of first n positive integers is given by ½ (n (n+1))
   Here n = 500
   Therefore sum = ½ (500(500+1)) = 125250.
7. 24
   (Explanation: Let the three numbers be (a-d), a, (a+d).
   Given (a-d) + a + (a+ d) = 72
   3a = 72
   a = 72/3 = 24.
8. 24
   (Explanation: Given n th term = 3n +2
   First term = 3 x 1 + 2 = 5
   Second term = 3 x 2 + 2 = 8
   Third term = 3 x 3 + 2 = 9 + 2 = 11
   Sum of first three terms = 5 + 8 + 11 = 24.
9. -32
   (Explanation: If we write the given AP in the reverse order,
   then a = -62 and d = 3.
   So 11 th term = -62 + (11 -1) 3 = -62 + 30 = -32.)
10. 13
    (Explanation: Let n th term = 0
    Here a = 84, d = -7
    n th term = a + (n-1) d = 0
    84 + (n-1) (-7) = 0
    84 – 7n + 7 = 0
    91 – 7n = 0
    91 = 7n
    n = 91/7 = 13.)