Let $S_n$,$S_{2n}$,and $S_{3n}$ be the sums of $n$,$2n$,and $3n$ terms of an Arithmetic Progression $(AP)$,respectively. Prove that $S_{3n} = 3(S_{2n} - S_n)$.

(N/A) Let the first term of the Arithmetic Progression be $a$ and the common difference be $d$.
The sum of $n$ terms is given by $S_n = \frac{n}{2}[2a + (n-1)d]$.
Similarly,$S_{2n} = \frac{2n}{2}[2a + (2n-1)d] = n[2a + (2n-1)d] = 2an + 2n^2d - nd$.
And $S_{3n} = \frac{3n}{2}[2a + (3n-1)d] = 3an + \frac{3n(3n-1)d}{2}$.
Now,consider the expression $3(S_{2n} - S_n)$:
$S_{2n} - S_n = n[2a + (2n-1)d] - \frac{n}{2}[2a + (n-1)d]$
$= \frac{n}{2} [2(2a + 2nd - d) - (2a + nd - d)]$
$= \frac{n}{2} [4a + 4nd - 2d - 2a - nd + d]$
$= \frac{n}{2} [2a + 3nd - d] = \frac{n}{2} [2a + (3n-1)d]$.
Multiplying this by $3$,we get $3(S_{2n} - S_n) = 3 \times \frac{n}{2} [2a + (3n-1)d] = \frac{3n}{2} [2a + (3n-1)d] = S_{3n}$.
Hence,$S_{3n} = 3(S_{2n} - S_n)$ is proved.