Solutions:

In the given AP. 9, 17, 25 … let there be n terms

For the above A.P.,

First term, a = 9

The common difference, d = a₂−a₁ = 17−9 = 8

Say that the total of n terms is;

S_n = n/2 [2a+(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 fraction or negative, As a result, n = 12 only.