How many terms of the AP 9, 17, 25, ... must be taken so that their sum is 636?

Here, first term = a = 9

Common difference = d = 17 - 9 = 8

Let first n terms of the AP sums to 636.

∴ S_n = 636

To find: n

Now, S_n = (n/2) × [2a + (n - 1)d]

Since, S_n = 636

∴ (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

⇒ 4n² + 5n - 636 = 0

⇒ 4n² + 5n - 636 = 0

⇒ (n - 12)(4n + 53) = 0

⇒ n = 12 or n = - 53/4

But n can’t be negative and fraction.

∴ n= 12

∴ 12 terms of the given AP sums to 636.