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 = ½

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