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 Sn = 636
Now
We know:
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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