Question

The sum of $7^{th}$ & $3^{rd}$ term of an AP is $6$ and their product is $8$. Find the sum of first $20$ terms of the $AP$.

Answer 1

$\Rightarrow$$a_3+a_7=6$

$\Rightarrow a+2d+a+6d=6$

$\therefore a+4d=3$

$\Rightarrow a=3-4d$...(1)

$\Rightarrow$$a_3\times a_7=8$

$\Rightarrow (a+2d)(a+6d)=8$...(2)

Substituting the value of (1) in (2), we get,

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

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

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

$\Rightarrow d=\pm \frac{1}{2}$

When $d=\frac{1}{2}\Rightarrow a=3-4\left(\frac{1}{2}\right)\Rightarrow a=1$

When $d=\frac{-1}{2}\Rightarrow a=3-4(-\frac{1}{2})\Rightarrow a=5$

$\Rightarrow$$S_{20}$ when $d=\frac{1}{2}$ and $a=1$

$\Rightarrow$$S_{20}=\frac{20}{2}\left[2\right(1)+(20-1\left)\frac{1}{2}\right]$

$=10[2+\frac{19}{2}]$

$=115$

$\Rightarrow$$S_{20}$ when $d=-\frac{1}{2}$ and $a=5$

$\Rightarrow$$S_{20}=\frac{20}{2}\left[2\right(5)+(20-1\left)(-\frac{1}{2})\right]$

$=10[10-\frac{19}{2}]$

$=5$