Question

If $S_n$ denotes the sum of the first $n$ terms of an AP, prove that $S_{12}=3(S_8-S_4)$

Answer 1

We have to prove: $S_{12}=3(S_8-S_4)$
Let $a$ is the first term of AP and $d$ is the common difference.

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

Now,

$S_{12}=\frac{12}{2}\left(2a+(12-1)d\right)=12a+66d$
$S_8=\frac{8}{2}\left(2a+7d\right)=8a+28d$
$S_4=\frac{4}{2}\left(2a+3d\right)=4a+6d$

$LHS=S_{12}=12a+66d$
$RHS=3(S_8-S_4)=3(8a+28d-4a-6d)=12a+66d$
$LHS=RHS$

Hence proved.