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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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Sol. Let first term $= a$ and common difference $= d$
ATQ, $(a + 2d) + (a + 6d) = 6$
$a + 4d = 3$
$a = 3-4d$
Also, $(a + 2d)(a + 6d) = 8$
$(3 - 4d + 2d)(3 - 4d + 6d) = 8$
$9 - 4d^2 = 8$
$d = \pm \frac{1}{2}$
When $d = \frac{1}{2} \Rightarrow a = 1$
$S_{16} = \frac{16}{2}[2 \times 1 + 15 \times \frac{1}{2}]$
$= 76$
When $d = -\frac{1}{2} \Rightarrow a = 5$
$S_{16} = \frac{16}{2}[2 \times 5 + 15 \times (-\frac{1}{2})]$
$= 20$
ATQ, $(a + 2d) + (a + 6d) = 6$
$a + 4d = 3$
$a = 3-4d$
Also, $(a + 2d)(a + 6d) = 8$
$(3 - 4d + 2d)(3 - 4d + 6d) = 8$
$9 - 4d^2 = 8$
$d = \pm \frac{1}{2}$
When $d = \frac{1}{2} \Rightarrow a = 1$
$S_{16} = \frac{16}{2}[2 \times 1 + 15 \times \frac{1}{2}]$
$= 76$
When $d = -\frac{1}{2} \Rightarrow a = 5$
$S_{16} = \frac{16}{2}[2 \times 5 + 15 \times (-\frac{1}{2})]$
$= 20$