Number of Sides of a Polygon with AP Angles

The sum of interior angles of an $n$-sided polygon is given by: $S = (n - 2) \times 180^\circ$

Also, using the formula for the sum of an arithmetic progression: $S = \frac{n}{2} [2a + (n - 1)d] = \frac{n}{2} [2(120^\circ) + (n - 1)5^\circ] = \frac{n(5n + 235)}{2}$

Equating both sums: $\frac{n(5n + 235)}{2} = (n - 2) \times 180$ $5n^2 + 235n = 360n - 720$ $5n^2 - 125n + 720 = 0$ $n^2 - 25n + 144 = 0$ $(n - 9)(n - 16) = 0$

So $n = 9$ or $n = 16$.

If $n = 16$, the largest angle would be $120^\circ + 15 \times 5^\circ = 195^\circ > 180^\circ$, which is impossible for a convex polygon. Hence, $n = 9$.