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.

From the given statements, we can write,
\(a_3 + a_7 = 6\) …………………………….(i)
And
\(a_3 ×a_7 = 8\) ……………………………..(ii)

By the \(n^{th}\) term formula,
\(a_n = a+(n - 1)d\)

Third term, \(a_3 = a+(3 -1)d\)
\(a_3 = a + 2d\)………………………………(iii)
And Seventh term, \(a_7= a+(7-1)d\)
\(a_7 = a + 6d \)………………………………..(iv)

From equation (iii) and (iv), putting in equation(i), we get,
\(\Rightarrow \) a+2d +a+6d = 6
\(\Rightarrow \) 2a+8d = 6
\(\Rightarrow \) a+4d=3
\(\Rightarrow \) a = 3–4d …………………………………(v)

Again putting the eq.(iii) and (iv), in eq. (ii), we get,
(a+2d)×(a+6d) = 8

Putting the value of a from equation (v), we get,
\(\Rightarrow (3–4d +2d)×(3–4d+6d) = 8\)
\(\Rightarrow (3 –2d)×(3+2d) = 8\)
\(\Rightarrow 3^2 – 2d^2\) = 8
\(\Rightarrow 9 – 4d^2\) = 8
\(\Rightarrow 4d^2 = 1\)
\(\Rightarrow d = {{1} \over {2}} or -{{1} \over {2}}\)

Now, by putting both the values of d, we get,
\(\Rightarrow a = 3 – 4d = 3 – 4({{1} \over {2}}) = 3 – 2 = 1\), when \(d = {{1} \over {2}}\)

\(a = 3 – 4d = 3 – 4(-{{1} \over {2}}) = 3+2 = 5\), when \(d = -{{1} \over {2}}\)

We know, the sum of nth term of AP is;
\(S_n = {{n} \over {2}} [2a +(n – 1)d]\)

So, when a = 1 and d=\({{1} \over {2}}\)

Then, the sum of first 16 terms are;
\(S_{16} = {{16} \over {2}} [2 +(16-1){{1} \over {2}}] \)
\(= 8(2+{{15} \over {2}}) = 76\)

And when a = 5 and d= -\({{1} \over {2}}\)
Then, the sum of first 16 terms are;

\(S_{16} = {{16} \over {2}} [2.5+(16-1)(-{{1} \over {2}})] \)
\( = 8({{5} \over {2}})=20\)