The sum of the third term and the seventh term 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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Step 1: Let first term $= a$ and common difference $= d$.
ATQ, $(a + 2d) + (a + 6d) = 6 \Rightarrow a + 4d = 3 \Rightarrow a = 3 - 4d$
Step 2: Also, $(a + 2d)(a + 6d) = 8$
$(3 - 4d + 2d)(3 - 4d + 6d) = 8$
$(3 - 2d)(3 + 2d) = 8$
$9 - 4d^2 = 8$
Step 3: $d = \pm\frac{1}{2}$
Step 4: When $d = \frac{1}{2} \Rightarrow a = 1$.
$S_{16} = \frac{16}{2}\left[2 	imes 1 + 15 	imes \frac{1}{2}ight] = 76$
Step 5: When $d = -\frac{1}{2} \Rightarrow a = 5$.
$S_{16} = \frac{16}{2}\left[2 	imes 5 + 15 	imes \left(-\frac{1}{2}ight)ight] = 20$

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