Question

The sum of n, 2n, 3n terms of an A.P. are $S_1$, $S_2$, $S_3$ respectively. Prove that $S_3$ = 3($S_2$ – $S_1$).

Answer 1

Let a be the first term and d be the common difference of the given A.P. Then,
$S_1$ = Sum of n terms

$S_1$ =($\frac{n}{2}$) {2a+(n— 1) d} ...... (i)
$S_2$ = Sum of 2n terms

$S_2$ = ($\frac{2n}{2}$) [2a + (2n — l) d] (ii)
and, $S_3$ = Sum of 3n terms

$S_3$ = ($\frac{3n}{2}$)[2a + (3n — l) d] (iii)

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

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

Therefore, 3($S_2$- $S_1$) = $\frac{3n}{2}$ [2a + (3n - l) d] = $S_3$ [Using (iii)]
Hence, $S_3$ = 3 ($S_2$ - $S_1$) ...[Using (iii)]