Let the sum of $n, 2n, 3n$ terms of an $A.P.$ be $S_{1}, S_{2}$ and $S_{3}$ respectively. Show that $S_{3} = 3(S_{2} - S_{1})$.

Let $a$ be the first term and $d$ be the common difference of the $A.P.$
$S_{1} = \frac{n}{2}[2a + (n - 1)d]$ $(1)$
$S_{2} = \frac{2n}{2}[2a + (2n - 1)d] = n[2a + (2n - 1)d]$ $(2)$
$S_{3} = \frac{3n}{2}[2a + (3n - 1)d]$ $(3)$
Now,calculate $S_{2} - S_{1}$:
$S_{2} - S_{1} = n[2a + (2n - 1)d] - \frac{n}{2}[2a + (n - 1)d]$
$= \frac{n}{2} [2(2a + 2nd - d) - (2a + nd - d)]$
$= \frac{n}{2} [4a + 4nd - 2d - 2a - nd + d]$
$= \frac{n}{2} [2a + 3nd - d] = \frac{n}{2} [2a + (3n - 1)d]$
Therefore,$3(S_{2} - S_{1}) = 3 \times \frac{n}{2} [2a + (3n - 1)d] = \frac{3n}{2} [2a + (3n - 1)d] = S_{3}$.
Hence,the result is proved.