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;
Sn = 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)
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
n cannot be negative or fraction, therefore, n = 12 only.
Therefore, the required number of terms = 12.