# Arithmetic Progressions – Class 10

**CBSE CLASS 10 MATHEMATICSArithmetic Progressions – Chapter 5 /MCQ**

- The 10 th term of the AP: 2, 7, 12……………. is —————-

A. 43

B. 45

C. 47 - 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 - Which term of the AP 1125, 1100, 1075, 1050 ……….. is zero?

A. 45

b. 46

c. 47 - In the following AP, find the missing term: 2, ———, 26

A. 12

B. 13

C. 14 - Which term of the AP: 4, 7, 10, 13 ………….. is 76?

A. 20

B. 25

C. 30 - The sum of first 500 positive integers is given by ———–

A. 125125

B. 125250

C. 125275 - If the sum of three numbers in AP is 72, then the middle term is ———–

A. 20

B. 24

C. 28 - 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 - 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 - Which term of the AP: 84, 77, 70, 63 ———– is zero?

A. 11

B. 12

C. 13

**Answers with explanation:**

- 47

(Explanation: Here a = 2, d=5, n=10

n th term = a + (n-1)d = 2 + (10 – 1) 5 = 2 + 45 = 47) - 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) - 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.) - 14

(Explanation: Middle term = ½ (26 +2) = 14.) - 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) - 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. - 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. - 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. - -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.) - 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.)