Question

If $S_n$ denotes the sum of the first $n$ terms of an A.P., prove that $S_{30}=3(S_{20}-S_{10})$

Answer 1

Let the First term of $A.P=a$

And common difference $=d$

$S_{30}=\frac{30}{2}[2a+(30-1\left)d\right]$

$S_{20}=\frac{20}{2}\left[2a\right(20-1\left)d\right]$

$S_{10}=\frac{10}{2}[2a+(10-1\left)d\right]$

$R.H.S=3[S_{20}-S_{10}]$

$=3[\left[\frac{20}{2}\right(2a+20-1\left)d\right]-\left[\frac{10}{2}\right(2a+(10-1)d]]$

$=3[20a+190d-10a-45d]$

$=15(2a+29d)$

$=\frac{30}{2}[2a+(30-1\left)d\right]$

$=S_{30}$

$=L.H.S$