4.031. It is known that the interior angles of a certain convex polygon, the smallest of which is $120^{\circ}$, form an arithmetic progression with a difference of $5^{\circ}$. Determine the number of sides of the polygon.
Solution.

From the condition, we have $a_{1}=120^{\circ}, d=5^{\circ}$. Using the formulas for the sum of terms of an arithmetic progression (4.5) and the sum of the interior angles of an $n$-sided polygon $S_{n}=180^{\circ}(n-2)$, we get

$$
\frac{240^{\circ}+(n-1) 5^{\circ}}{2} \cdot n=180^{\circ}(n-2), n^{2}-25 n+144=0 \Rightarrow
$$

$\Rightarrow n_{1}=9, n_{2}=16$ (the latter does not fit, as in this case

$\Rightarrow a_{16}=120^{\circ}+5^{\circ} \cdot 15=195^{\circ}$, while the interior angle of a convex $n$-sided polygon is always less than $\left.180^{\circ}\right)$.

Answer: 9.