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.

To Find: S₁₆ 
Given: a₃ + a₇ = 6
           a₃ x a₇ = 8

nth term of an AP is given by the formula
a_n = a + (n - 1) d
where, a_n = nth term 
n = number of term
d = common difference
So now its given that sum of third and seventh term is 6, thus we need to find 3rd and 7th term first,
a₃ = a + 2d

a₇ = a + 6d

As per question;

a₃ + a₇ = 6

So now,

a + 2d + a + 6d = 6

2a + 8d = 6

a + 4d = 3

a = 3 – 4d    ....................eq (i)

Similarly,

Product of third and seventh term is given as 8. So,

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

a² + 6ad + 2ad + 12d² = 8  ..........................eq(ii)

Substituting the value of a in equation (ii), 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:

S₁₆ =  8[ 2 + (15/2)]

= 4 x 19

S₁₆ = 76

Thus, sum of first 16 terms of this AP is 76.
Now by taking d = -1/2, we get,
a = 3 - 4 (-1/2)
a = 3 + 2 = 5

S = 8[10 - 15/2]
S = 4[20 - 15]
S = 4[5] = 20
So, another possible value of sum is 20.