Question about interior angles and number of sides is unclear

Question

The degree measure of the smallest interior angle of an $n$-sided polygon is 120°. The other angles each have an integral degree measure that is 5° more than the previous angle. Find all possible values of $n$.

I have never encountered a question similar to this. I am a highschool student and have already taken geometry but I don't know that this question defines the other angles of the polygon to be, specifically in the bold area.

Answer 1

Evidently it means that one of the angles is $120^\circ$ and other interior angles are increasing by $5^\circ$ if we go clockwise or anti-clockwise. Obviously there will be a decrease if we come all the way back from last angle to the angle we started with!

You have that for an $n$ sided polygon, sum of interior angles of a polygon is given by $$(n-2)\cdot 180^\circ$$

Now the rest is summation of interior angles which are in $AP$. Here I take all angles in degrees

$$ \frac{n}{2} \left(2\cdot 120 + (n-1)\cdot 5\right) = (n-2)\cdot 180$$

This gives two values of $n$, namely $n=9, 16$ and I think only one value holds, because other might give angles $> 180^\circ$.

Answer 2

Note that the angle sum of a polygon with $n$ sides is $180(n-2)$ degrees. The angel measures are $ 120, 125,130,..., 120+5(n-1)$ which is an arithmetic sequence with $n$ terms. The sum of these terms is $ n(235+5n)/2.$ Therefore we get $ n(235+5n)/2 = 180(n-2)$ . The resulting quadratic equation $ n^2-25n+144=0$ has two solutions $n=9$ and $n=16.$

Answer 3

So the exterior angles are $60^\circ,55^\circ,50^\circ,\ldots$ etc. The exterior angles of a convex polygon add up to $360^\circ$, so you need to find how many terms in the sum $60+55+50+\cdots$ you need to reach $360$.