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