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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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