Question

Let sum of n, 2n, 3n terms of an A.P. be $S_1,S_2$ and $S_3$ respectively, show that $S_3=3(S_2-S_1)$

Answer 1

Let 'a' be the 1st term and 'd' be the common difference of the given A.P.

$\therefore S_n=\frac{n}{2}[2a+(n-1\left)d\right]$ ..... (1)

$\therefore S_2=\frac{2n}{2}[2a+(2n-1\left)d\right]$ ..... (2)

$\therefore S_3=\frac{3n}{2}[2a+(3n-1\left)d\right]$ ..... (3)

Now $S_2-S_1=\frac{2n}{2}[2a+(2n-1\left)d\right]-\frac{n}{2}[2a+(n-1\left)d\right]$

$=\frac{n}{2}[4a+4nd-2d-2a-nd+d]$

$=\frac{n}{2}[2a+3nd-d]$

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

$\therefore 3(S_2-S_1)=\frac{3n}{2}[2a+(3n-1\left)d\right]=S_3[\because S_3=\frac{3n}{2}[2a+(3n-1)d\left]\right]$

Thus, $S_3=3(S_2-S_1)$