Question

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

Answer 1

Let there be n terms of this A.P.

For this A.P., a = 9

d = a₂ − a₁ = 17 − 9 = 8

636 = n [9 + 4n − 4]

636 = n (4n + 5)

4n² + 5n − 636 = 0

4n² + 53n − 48n − 636 = 0

n (4n + 53) − 12 (4n + 53) = 0

(4n + 53) (n − 12) = 0

Either 4n + 53 = 0 or n − 12 = 0

or n = 12

n cannot be . As the number of terms can neither be negative nor fractional, therefore, n = 12 only.