Question

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

Answer 1

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