If sn denotes the sum of first


If sn denotes the sum of first n terms of an AP, then prove that   s12 =3(s8-s4)


$\because$ Sum of $n$ terms of an AP, $S_{n}=\frac{n}{2}[2 a+(n-1) d]$ ...(i)

 $\therefore$ $S_{8}=\frac{8}{2}[2 a+(8-1) d]=4(2 a+7 d)=8 a+28 d$ 

and $S_{4}=\frac{4}{2}[2 a+(4-1) d]=2(2 a+3 d)=4 a+6 d$

Now, $S_{8}-S_{4}=8 a+28 d-4 a-6 d=4 a+22 d$ ... (ii)

and   $S_{12}=\frac{12}{2}[2 a+(12-1) d]=6(2 a+11 d)$

$=3(4 a+22 d)=3\left(S_{8}-S_{4}\right) \quad$ [from Eq. (ii)]

$\therefore$ $S_{12}=3\left(S_{8}-S_{4}\right)$ Hence proved.


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