The sum of the third and the seventh termsof an AP is 6 and their product is 8. Findthe sum of first sixteen terms of the AP.

a3 = a + 2d


a7 = a + 6d


As per question;


a3 + a8 = 6


a + 2d + a + 6d = 6


2a + 8d = 6


a + 4d = 3


a = 3 – 4d (i)


Similarly,


(a + 2d)(a + 6d) = 8


a2 + 6ad + 2ad + 12d2 = 8


Substituting the value of a in equation (i), we get;


(3 – 4d)2 + 8(3 – 4d)d + 12d2 = 8


9 – 24d + 16d2 + 24d – 32d2 + 12d2 = 8


9 – 4d2 = 8


2d = 1


d = 1/2


Using the value of d in equation (1), we get;


a = 3 – 4d


Or, a = 3 – 2 = 1


Sum of first 16 terms is calculated as follows:




= 8[ 2 + (15/2)]


=4*19


= 76


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