The sum of the third and seventh term of an A.P. is 6 and their product is 8. Find the sum of the first sixteen terms of the A.P. .

Let *a* and *d* be the first term and common difference of A.P.

*n*^{th} term of A.P., *a _{n}* =

*a*+ (

*n*– 1)

*d*

∴ *a*_{3} = *a* + (3 – 1) *d* = *a* + 2*d*

*a*_{7} = *a* + (7 – 1) *d* = *a* + 6*d*

Given, *a*_{3} + *a*_{7} = 6

∴ (*a* + 2*d*) + (*a* + 6*d*) = 6

⇒ 2*a* + 8*d* = 6

⇒ *a* + 4*d* = 3 ...(1)

Given, *a*_{3} × *a*_{7} = 8

∴ (*a* + 2*d*) + (*a* + 6*d*) = 8

⇒ (3 – 4*d* + 2*d*) (3 – 4*d* + 6*d*) = 8 [ Using (1) ]

⇒ (3 – 2*d*) (3 + 2*d*) = 8

⇒ 9 – 4*d*^{2}* *= 8

⇒ 4*d*^{2}* *= 1

*a* = 3 – 4*d* = 3 – 2 = 1

When *a* = 1 and

Thus, the sum of first 16 terms of the A.P. is 76 or 20.

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