Task B-4.2. Determine the number of sides of a polygon for which the measure of the smallest angle is $132^{\circ}$, and each subsequent angle is $2^{\circ}$ larger.
## Solution.

The measures of the angles form an arithmetic sequence with $a_{1}=132^{\circ}, d=2$.

The sum of $n$ terms of the arithmetic sequence is $S=\frac{n}{2}\left(2 a_{1}+(n-1) d\right)=n(131+n)$.

The sum of all angles in a polygon is $S=(n-2) \cdot 180$.

Then from $n(131+n)=(n-2) \cdot 180$, it follows that $n^{2}-49 n+360=0$.

The solutions to this quadratic equation are $n_{1}=9, n_{2}=40$. Therefore, the polygon can have 9 or 40 sides. In a polygon with 40 sides, one angle will be $180^{\circ}$ due to the conditions of the problem, so such a polygon is not a solution.

(2 points)