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

Given, a = 9, d = 17-9= 8 , S_n = 636

Using the formula,

S_n = n/2 [2a + (n - 1)d]

636 = n/2 [2 × 9 + (n - 1)8]

636 = n/2 [18 + 8n - 8]

636 = n/2 [10 + 8n]

636 = n[5 + 4n]

636 = 5n + 4n²

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

n = - 53/4 or n =12

The number of terms can not be negative.

Therefore, n = 12