Question

The sum of the third and seventh term of an A.P. is $6$ and their product is $8$. Find the sum of first sixteen terms of the A.P.

Answer 1

Let $a$ be the common difference of the A.P.
We have,
$a_3+a_7=6$ and $a_3{a}_7=8$

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

$⟹2a+8d=6$ and $(a+2d)(a+6d)=8$

$⟹a+4d=3$ and $(a+2d)(a+6d)=8$

$⟹a=3-4d$ and $(a+2d)(a+6d)=8$ [Putting $a=3-4d$ in the second equation]

$⟹(3-4d+2d)(3-4d+6d)=8$

$⟹9-4d^2=8⟹4d^2=1⟹d^2=\frac{1}{4}⟹d=\pm \frac{1}{2}$

Case $I$ When $d=\frac{1}{2}:$
Putting $d=\frac{1}{2}$ in $a=3-4d$, we get
$a=3-4\times \frac{1}{2}=3-2=1$
$\therefore S_{16}=\frac{16}{2}\{2a+(16-1\left)d\right\}=8\{2\times 1+15\times \frac{1}{2}\}=8\times \frac{19}{2}=76$
Case $II$ When $d=-\frac{1}{2}:$
Putting $d=-\frac{1}{2}$ in $a=3-4d$, we get $a=3+2=5$
$S_{16}=\frac{16}{2}\{2a+(16-1\left)d\right\}$
$⟹S_{16}=8\{10+15\times -\frac{1}{2}\}=8\times \frac{5}{2}=20$