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

Given that, a = 9, d = 8 and S_n = 636

Now

We know:

636

636= (18 + 8N -8)

636 = N(9 +4N -4)

4N² +5N -636 =0

2N² -N -45 =0

(4N +53)(N – 12)

Now, n = - 53/4 and n = 12

Taking the integral value and rejecting the fractional value, we have n = 12