Explanation

(c)
Let the number of sides of the polygon be $n$.Then, the sum of interior angles of the polygon
$= (2n - 4)\frac{\pi}{4} = (n - 2)\pi$
Since, the angles are in AP and $a = 120{^\circ},\ d = 5$
Therefore, $S_{n} = \frac{n}{2}\lbrack 2a + (n - 1)d\rbrack$
$\Rightarrow \frac{n}{2}\left\lbrack 2 \times 120 + (n - 1)5 \right\rbrack = (n - 2)180$
$\Rightarrow n^{2} - 25n + 144 = 0$
$\Rightarrow (n - 9)(n - 16) = 0$
$\Rightarrow n = 9,\ 16$
Take $n = 16$
$T_{16} = a + 15d = 120{^\circ} + 15(5{^\circ}) = 195{^\circ}$, which is impossible, an interior angle cannot be greater than $180{^\circ}$.
Hence, $n = 9$