Question

Question 26
If $S_n$ denotes the sum of first n terms of an AP, then prove that $S_{12}=3(S_8-S_4)$

Answer 1

$\because$ Sum of n terms of an AP, $S_n=\frac{n}{2}[2a+(n-1\left)d\right]\cdots \left(i\right)$
$\therefore S_8=\frac{8}{2}[2a+(8-1\left)d\right]=4(2a+7d)=8a+28d$
$And{S}_4=\frac{4}{2}[2a+(4-1\left)d\right]=2(2a+3d)=4a+6d$
$Now,S_8-S_4=8a+28d-4a-6d=4a+22d\cdots \left(ii\right)$
$And{S}_{12}=\frac{12}{2}[2a+(12-1\left)d\right]=6(2a+11d)$
$=3(4a+22d)=3(S_8-S_4)[fromEq.(ii\left)\right]\therefore S_{12}=3(S_8-S_4)$