The interior angles of a polygon are in A.P. The smallest angle is $120^{\circ}$ and the common diff

Question

The interior angles of a polygon are in A.P. The smallest angle is $120^{\circ}$ and the common difference is $5^{\circ}$. Find the number of sides of the polygon. \

Professor PC Thomas Classes · Answer

Let the number of sides of the polygon be $n$\ Then the sum of all the interior angles $=\left(n 	imes 180^{\circ}-360^{\circ}ight)$\ Sum of the interior angles\ $=120+125+130+\ldots$ to $n$ terms\ $=\frac{n}{2}[240+(n-1) 5]$\ $	herefore \frac{n}{2}[240+(n-1) 5]=n 	imes 180-360$\ $\Rightarrow n^{2}-25 n+144=0$\ $\Rightarrow(n-9)(n-16)=0 \Rightarrow n=9$ or 16\ But when $n=16$, the greatest interior angle is $120^{\circ}+(16-1) 5=195^{\circ}$ which is not possible, for interior angle is < 180 .\ Hence the number of sides $=9$.