As the sum of first $ n $ terms of an AP is denoted by $ \mathrm{S}_{n} $, prove that $ \mathrm{S}_{12}=3\left(\mathrm{~S}_{8}-\mathrm{S}_{4}\right) $ for any AP.

Given: 

The sum of first \( n \) terms of an AP is denoted by \( \mathrm{S}_{n} \).

To do: 

We have to prove that \( \mathrm{S}_{12}=3\left(\mathrm{~S}_{8}-\mathrm{S}_{4}\right) \) for any AP.

Solution:

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

We know that,

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

This implies,

$\mathrm{S}_{12}=\frac{12}{2}[2 a+(12-1) d]$

$\mathrm{S}_{12}=6[2 a+11 d]$

$\mathrm{S}_{12}=12 a+66 d$.........(i)

$\mathrm{S}_{8}=\frac{8}{2}[2 a+(8-1) d]$

$\mathrm{S}_{8}=4[2 a+7 d]$

$\mathrm{S}_{8}=8 a+28 d$............(ii)

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

$\mathrm{S}_{4}=2[2 a+3 d]$

$S_{4}=4 a+6 d$.........(iii)

Therefore,

$3(\mathrm{~S}_{8}-\mathrm{S}_{4})=3(8 a+28 d-4 a-6 d)$

$=3(4 a+22 d)$

$=12 a+66 d$

Hence, $3(S_{8}-S_{4})=S_{12}$.

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Updated on: 10-Oct-2022

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