Q4 of 49 Page 112

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:



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


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