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

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

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

⇒ 636 = 9n + 4n² − 4n

⇒ 4n² + 5n − 636 = 0

⇒ 4n² + 53n − 48n − 636 = 0

⇒ n(4n + 53) − 12(4n + 53) = 0

⇒ (4n + 53) (n − 12) = 0

⇒ 4n + 53 = 0 or n − 12 = 0

⇒ n = `(-53)/4` or n = 12

As the number of terms can neither be negative nor fractional, therefore, n = 12 only.