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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 first sixteen terms of the AP?

Given:

The sum of the third term and the seventh term of an AP is 6 and their product is 8.

To do:

We have to find the sum of first sixteen terms of the AP.

Solution:

Let $a$ and $d$ be the first term and the common difference of an AP.

This implies,

The AP formed is $a, (a+d), (a+2 d), \ldots$

$a_{3}+a_{7}=6$

$(a+2 d)+(a+6 d)=6$

$2a+8d=6$

$a+4d=3$

$a=3-4d$

$a_{3} \times a_{7}=8$

$(a+2 d)(a+6 d)=8$

$[(3-4 d)+2 d][(3-4 d)+6 d]=8$

$(3-2 d)(3+2 d)=8$

$9-4 d^{2}=8$

$d^{2}=\frac{1}{4}$

$d^2=(\frac{1}{2})^{2}$

$d=\pm \frac{1}{2}$

If $d=\frac{1}{2}, a=3-4 \times \frac{1}{2}=1$

Therefore,

$S_{16}=\frac{16}{2}[2 \times 1+(16-1) \times \frac{1}{2}]$

$=8[2+\frac{15}{2}]$

$=8[\frac{19}{2}]$

$=76$

If $d=-\frac{1}{2}, a=3-4 \times (-\frac{1}{2})=3+2=5$

Therefore,

$S_{16}=\frac{16}{2}[2 \times 5+(16-1) \times \frac{-1}{2}]$

$=8[10+(\frac{-15}{2})]$

$=8[10-\frac{15}{2}]$

$=8 \times \frac{5}{2}$

$=20$

**Hence, the sum of first sixteen terms of the given AP is 20 or 76.**

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