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# 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;

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