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10th CBSE math solution for exercise 5.3 part 4

(vi) Given a= 2 , d = 8,S n = 90 find n and an

Solution:

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

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

90 x 2 = n [4 + 8 n - 8]

180 = n [-4 + 8 n]

180 = -4 n + 8 n²

8 n² - 4 n - 180 = 0

Divide the whole equation by 4

2 n² - n - 45 = 0

(n - 5) (2 n + 9) = 0

n = 5 or n = -9/2

number of terms may not be negative or fraction.

a₅ = a + 4 d

= 2 + 4 (8)

= 2 + 32

= 34

(vii) Given a= 8 , a n = 62 ,S n = 210 find n and d