Question:
If sn denotes the sum of first n terms of an AP, then prove that s12 =3(s8-s4)
Solution:
$\because$ 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.