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(C) We know that the $n^{th}$ term of an $AP$ is $a_n = a + (n-1)d$.Thus,$a_3 = a + 2d$ and $a_7 = a + 6d$.Given $a_3 + a_7 = 6$,we have $(a + 2d) + (

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$20$

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(C) We know that the $n^{th}$ term of an $AP$ is $a_n = a + (n-1)d$.Thus,$a_3 = a + 2d$ and $a_7 = a + 6d$.Given $a_3 + a_7 = 6$,we have $(a + 2d) + (a + 6d) = 6$,which simplifies to $2a + 8d = 6$,or $a + 4d = 3$. So,$a = 3 - 4d$ $...(i)$.Also,given $a_3 	imes a_7 = 8$,we have $(a + 2d)(a + 6d) = 8$.Substituting $a = 3 - 4d$ from $(i)$ into this equation:$(3 - 4d + 2d)(3 - 4d + 6d) = 8$$(3 - 2d)(3 + 2d) = 8$$9 - 4d^2 = 8 \implies 4d^2 = 1 \implies d^2 = 1/4 \implies d = \pm 1/2$.Case $1$: If $d = 1/2$,then $a = 3 - 4(1/2) = 3 - 2 = 1$.The sum of the first $16$ terms is $S_{16} = \frac{16}{2}[2(1) + (16-1)(1/2)] = 8[2 + 7.5] = 8[9.5] = 76$.Case $2$: If $d = -1/2$,then $a = 3 - 4(-1/2) = 3 + 2 = 5$.The sum of the first $16$ terms is $S_{16} = \frac{16}{2}[2(5) + (16-1)(-1/2)] = 8[10 - 7.5] = 8[2.5] = 20$.Since $20$ is one of the options,the correct answer is $20$.

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

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