# 10th CBSE math solution for exercise 5.3 part 5

This page 10th CBSE math solution for exercise 5.3 part 5 is going to provide you solution for every problems that you find in the exercise no 5.3 part 5

## 10th CBSE math solution for exercise 5.3 part 5 (ix) Given a = 3 , n = 8 ,S n = 192 find d

Solution:

S n = (n/2) [2a + (n - 1) d]

192 = (8/2) [2 (3) + (8 - 1) d]

192 = 4 [6 + 7 d]

192/4 = [6 + 7 d]

48 - 6 = 7 d

42/7 = d

d = 6

(x) Given l = 28 , S = 144 , and there are total 9 terms. Find a

Solution:

S n = (n/2) [a + l]

n = 9

144 = (9/2) [ a + 28]

(144 x 2)/9 = a + 28

32 =a + 28

a = 32 - 28

a = 4

In the page 10th CBSE math solution for exercise 5.3 part 5 next we are going to see the solution of next problem.

(4) How many terms of the AP 9,17,25,.......... must be taken to give a sum of 636?

Solution:

a n = a + (n - 1) d

S n = 636

S n = (n/2) [2a + (n - 1) d]

a = 7   d = 17 - 9    Sn = 636

= 8

636 = (n/2)[2(9) + (n-1) 8]

(636 x 2) = n[18 + 8 n - 8]

(636 x 2) = n[10 + 8 n]

636 = n[5 + 4n]

636 = 5n + 4 n²

4 n² + 5 n - 636 = 0

(4n + 53)(n - 12) = 0

n = -53/4         n = 12

Therefore 12 terms are need to give a sum of 636.

(5) The first term of an AP is 5, the last term is 45 and the sum is 400. Find the number of terms and common difference.

Solution:

a = 5          l = 45          Sn = 400

S n = (n/2)[a + l]

400 = (n/2)[5 + 45]

400 = (n/2)

400 = n(25)

400/25 = n

n = 16

Sn = (n/2)[2 a + (n- 1) d]

400 = (16/2) [2(5) + (16-1)d]

400 = 8 [10 + 15 d]

400/8 = 10 + 15 d

50 = 10 + 15 d

50 - 10 = 15 d

40 = 15 d

40/15 = d

d = 8/3