If the sum of first $7$ terms of an $AP$ is $49$ and that of $17$ terms is $289,$ find the sum of first $n$ terms.

Given that,
$S_{7} = 49$
$S_{17} = 289$
Using the formula for the sum of the first $n$ terms of an $AP$: $S_{n} = \frac{n}{2}[2a + (n - 1)d]$
For $n = 7$:
$S_{7} = \frac{7}{2}[2a + (7 - 1)d] = 49$
$\frac{7}{2}(2a + 6d) = 49$
$7(a + 3d) = 49$
$a + 3d = 7$ $...(i)$
For $n = 17$:
$S_{17} = \frac{17}{2}[2a + (17 - 1)d] = 289$
$\frac{17}{2}(2a + 16d) = 289$
$17(a + 8d) = 289$
$a + 8d = 17$ $...(ii)$
Subtracting equation $(i)$ from equation $(ii)$:
$(a + 8d) - (a + 3d) = 17 - 7$
$5d = 10$
$d = 2$
Substituting $d = 2$ in equation $(i)$:
$a + 3(2) = 7$
$a + 6 = 7$
$a = 1$
Now,calculating the sum of the first $n$ terms:
$S_{n} = \frac{n}{2}[2a + (n - 1)d]$
$S_{n} = \frac{n}{2}[2(1) + (n - 1)(2)]$
$S_{n} = \frac{n}{2}[2 + 2n - 2]$
$S_{n} = \frac{n}{2}(2n)$
$S_{n} = n^{2}$