If $ S_{n} $ denotes the sum of first $ n $ terms of an AP, prove that
$ S_{12}=3\left(S_{8}-S_{4}\right) $

Given:

$S_n$ denotes the sum of first $n$ terms of an A.P.

To do:

We have to prove that $S_{12} = 3(S_8 – S_4)$.

Solution:

Let $a$ be the first term and $d$ the common difference.

We know that,

$\mathrm{S}_{n}=\frac{n}{2}[2 a+(n-1) d]$

$S_{4}=\frac{4}{2}[2 \times a+(4-1) \times d]$

$=2[2a+3d]$

$=4a+6d$......(i)

$S_{8}=\frac{8}{2}[2 \times a+(8-1) \times d]$

$=4[2a+7d]$

$=8a+28d$......(ii)

$S_{12}=\frac{12}{2}[2 \times a+(12-1) \times d]$

$=6[2a+11d]$

$=12a+66d$......(iii)

From (i) and (ii)

$3(S_8-S_4)=3[8a+28d-(4a+6d)]$

$=3(8a+28d-4a-6d)$

$=3(4a+22d)$

$=12a+66d$

$=S_{12}$       (From (iii))

Hence proved.