Question

The sum of the third and seventth term of an AP is 6 and their product is 8. Find the sum of the first 16 terms of the A.P.

Answer 1

$T_3+T_7=6$

$a+2d+a+6d=6$

$2a+8d=6$

$a+4d=3$

$\Rightarrow a=3-4d$

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

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

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

$\Rightarrow (3{)}^2-(2d{)}^2=8$

$\Rightarrow 9-4d^2=8$

$\Rightarrow 1=4d^2$

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

$a=s-4\left(\frac{1}{2}\right)=1$

$S_{16}=\frac{16}{2}[2a+15d]$

$=8\left(2\right(1)+15\left(\frac{1}{2}\right))$

$=8(2+\frac{15}{2})$

$=8\times \frac{19}{2}=76$.