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

#### Solution

We know that,

*a*_{n} = *a* + (*n *− 1) *d*

*a*_{3} = *a* + (3 − 1) *d*

*a*_{3} = *a* + 2*d*

Similarly, *a*_{7} = *a* + 6*d*

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

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

2*a* + 8*d* = 6

*a* + 4*d* = 3

*a* = 3 − 4*d* (*i*)

Also, it is given that (*a*_{3}) × (*a*_{7}) = 8

(*a* + 2*d*) × (*a *+ 6*d*) = 8

From equation (*i*),

(3-4d+2d) x (3 - 4d + 6d) = 8

(3 - 2d) x (3 + 2d) = 8

9 - 4d^{2} = 8

4d^{2} = 9 - 8 =1

d^{2} = 1/4

d = ± 1/2

d = 1/2 or 1/2

From the equation (i)

(When d os 1/2)

a = 3- 4d

a = 3 -4(1/4)

= 3 - 2 = 1

(When d is - 1/2)

a = 3 - 4(-1/2)

a = 3+2 = 5

`S_n = n/2 [2a(n-1)s]`

(when a is 1 and d is 1/2 )

`S_16 = 16/2[2(1)+(16+1)(1/2)]`

= 8[2+15/2]

= 4(19) =76

(When a is 5 and d is -1/2)

`S_16=16/2[2(5)+(16-1)(-1/2)]`

= 8[10+(15)(-1/2)]

= 8(5/2)

= 20