Polygon with increasing angles

Let there be $n$ sides of the polygon. Then the sum of its interior angles is given by

$S_{n} = (n-2) \times 180^\circ$ .

Also, the sum of all the terms of an arithmetic progression is given by

$S_{n} = \frac{n}{2}[2a_{1}+ (n-1)d]$ where $a_{1}=120^\circ$ and $d=5^\circ$

Equating both the equations:

$(n-2) \times 180^\circ = \frac{n}{2}[2a_{1} + (n-1)d]$

On solving, we will get two values for $n \Rightarrow 16, 9$

But if the polygon is of 16 sides, the $13^{th}$ angle will be $180^\circ$ , which is not allowed.

Hence, the answer is $\Rightarrow \boxed{9}$

The sum of all external angles is 360.So 60 + 55 +50 +...=360. SUM of PA >>> (60 + 60 -5(n-1))n/2=360>>>solving this we find 2 answers: 16 and 9.Calculating the last angle of the 16th side polygon (60-16*5= -20) we see that he cannot exist because he is negative , so the answer is just the 9