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

(N/A) Let the first term of the $A.P.$ be $a$ and the common difference be $d$.
The sum of $n$ terms of an $A.P.$ is given by $S_n = \frac{n}{2}[2a + (n-1)d]$.
Given:
$S_1 = \frac{n}{2}[2a + (n-1)d]$
$S_2 = \frac{2n}{2}[2a + (2n-1)d] = n[2a + (2n-1)d]$
$S_3 = \frac{3n}{2}[2a + (3n-1)d]$
Now,consider the expression $3(S_2 - S_1)$:
$S_2 - S_1 = n[2a + (2n-1)d] - \frac{n}{2}[2a + (n-1)d]$
$= \frac{n}{2} [2(2a + 2nd - d) - (2a + nd - d)]$
$= \frac{n}{2} [4a + 4nd - 2d - 2a - nd + d]$
$= \frac{n}{2} [2a + 3nd - d]$
$= \frac{n}{2} [2a + (3n-1)d]$
Multiplying by $3$:
$3(S_2 - S_1) = 3 \times \frac{n}{2} [2a + (3n-1)d] = \frac{3n}{2} [2a + (3n-1)d] = S_3$.
Hence,$S_3 = 3(S_2 - S_1)$ is proved.