For an arithmetic sequence, $S=\frac{n}{2}(\text{first}+\text{last})$.

Set your arithmetic-series sum equal to $180(n-2)$ and solve for $n$. W‌ritten b‍y Anikο

Check that the largest interior angle is less than $180^\circ$ to rule out any invalid solutions.

In Desmos, let $x$ represent $n$. Enter:

• $y=\frac{x}{2}(5x+235)$ (arithmetic-series sum)
• $y=180(x-2)$ (polygon interior-angle sum)

Click the intersection(s) of the two graphs to get the $x$-value(s) where the sums match.

Desmos will show intersections at $x=9$ and $x=16$. Since a convex polygon must have all interior angles less than $180^\circ$, $n=16$ is invalid (it would make the largest angle $195^\circ$). Therefore, the number of sides is 9. Writ⁠t⁠еn‍ b‌у Аn‍ікο

Let $n$ be the number of sides (and interior angles). The angles are

So the last angle is $120+5(n-1)$ degrees.

For an arithmetic sequence,

Here,

A polygon with $n$ sides has interior-angle sum $180(n-2)$, so

Multiply by 2 and simplify:

So $n=9$ or $n=16$. (​Аnікο.а‌і)

A convex polygon must have every interior angle less than $180^\circ$.

• If $n=16$, the largest angle would be $120+5(16-1)=195^\circ$, which is not possible for a convex polygon.

Therefore, the polygon has 9 sides.