Question

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

Answer 1

Let $1^{st}$ term of $A.P=a$

and common difference $=d$

We have to prove that

$S_{12}=3(S_8-S_4)$

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

$S_8=\frac{8}{2}[2a+(8-1\left)d\right]$

$S_4=\frac{4}{2}[2a+(4-1\left)d\right]$

Now, $R.H.S=3(S_8-S_4)$

$=3\left[\frac{8}{2}\right(2a+(8-1)d-\frac{4}{2}(2a+(4-1\left)d\right]$

$=3[8x+28d-4a-6d]$

$=6(2a+11d)$

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

$=S_{12}$

$=L.H.S$