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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.