**Solution:**

Let there be n terms of the AP. 9, 17, 25 …

For this A.P.,

First term, a = 9

Common difference, d = a _{2}− a _{1} = 17−9 = 8

As, the sum of n terms is;

S_{n} = n /2 [2 a + (n – 1) d ]

636 = n /2 [2 × a + (8 – 1) × 8]

636 = n /2 [18 +( n – 1) × 8]

636 = n [9 + 4n − 4]

636 = n (4n + 5)

4n^{2} + 5n − 636 = 0

4n^{2} + 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

n cannot be negative or fraction, therefore, n = 12 only.

Therefore, the required number of terms = 12.