geometry

I think you will find this useful:

https://web2.0calc.com/questions/polygon_6

I got a different answer from Cphill...

We know sum of all the interior angles of a polygon having n sides = (n-2)180

Given smallest angle = 120

a = 120

d = 5

Sum, S = (n/2)(2a+(n-1)d)

(n/2)(240+(n-1)5) = (n-2)180

n(240+5n-5) = (n-2)360

5n2+235n-360n+720 = 0

5n2-125n+720 = 0

n2-25n+144 = 0

(n-9)(n-16) = 0

n = 9 or 16.

n = 16 cannot be possible since interior angle cannot be greater than 180.

(n-2)180  = sum of the progression  =  n/2 ( 120 + 120 + 5n - 5)

360n - 720 = 235n + 5 n^2

0 = 5 n^2 - 125 n + 720              

           Now use the quadratic formula to find   n = 9   or 16   ( 16 is too large for a convex polygon....the angle would be 200^o)