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

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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.