Question

The sum of the third and the ninth term of an arithmetic progression is $8$. Find the sum of the first $11$ terms of the progression.

Answer 1

Sum of $3^{rd}$ and $9^{th}$ term of an $AP$ series $=8$

Let first term $=a$ and common difference$=d$

$a_r=a+(r-1)d$ [$r^{th}$ term]

$a+2d+a+8d=82a+10d=8-\left(i\right)$

Sum of first $11$ terms.

$S=\frac{n}{2}(2a+(n-1\left)d\right)=\frac{11}{2}(2a+10d)$

$=\frac{11}{2}\left(8\right)$ [As from $\left(i\right)2a+10d=8$]

$=44$

Sum of first $11$ terms $=44$