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

Q : 2    The sum of the third and the seventh terms of an AP is $\small 6$ and their product is $\small 8$. Find the sum of first sixteen terms of the AP.

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It is given that  sum of the third and the seventh terms of an AP is $\small 6$ and their product is $\small 8$
$a_3= a+ 2d$
$a_7= a+ 6d$
Now,
$a_3+a_7= a+ 2d+a+6d= 6$
$\Rightarrow 2a+8d = 6$
$\Rightarrow a+4d = 3 \Rightarrow a = 3-4d \ \ \ \ \ \ \ \ \ \ \ \ -(i)$
And
$a_3.a_7 = (a+2d).(a+6d)=a^2+8ad +12d^2 = 8 \ \ \ \ \ \ \ -(ii)$
put value from equation (i) in (ii) we will get
$\Rightarrow (3-4d)^2+8(3-4d)d+12d^2= 8$
$\Rightarrow 9+16d^2-24d+24d-32d^2+12d^2=8$
$\Rightarrow 4d^2 = 1$
$\Rightarrow d = \pm \frac{1}{2}$
Now,
case (i)  $d = \frac{1}{2}$

$a= 3 - 4 \times \frac{1}{2} = 1$
Then,
$S_{16}=\frac{16}{2}\left \{ 2\times 1+(16-1)\frac{1}{2} \right \}$

$S_{16}=76$

case (ii)  $d = -\frac{1}{2}$
$a= 3 - 4 \times \left ( -\frac{1}{2} \right ) = 5$

Then,
$S_{16}=\frac{16}{2}\left \{ 2\times 1+(16-1)\left ( -\frac{1}{2} \right ) \right \}$

$S_{16}=20$

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