Question

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

Answer 1

Sum of the interior angles of a polygon of n sides.
$=(2n-4)\pi /2=(n-2)\pi =(n-2)\cdot {180}^o$.
Also $a={120}^o,d=5^o$.
$\therefore \frac{n}{2}[2\cdot {120}^o+(n-1\left)5^o\right]=(n-2)\cdot {180}^o$
Cancel $5$ and simplify.
$\therefore n^2-25n+144=0$
$\therefore (n-9)(n-16)=0$.
$\therefore n=9,16$.
But when $n=16,T_n=a+(16-1)d$
$={120}^o+15\times 5^o={195}^o$
This is not possible as interior angle cannot be greater than ${180}^o$,
$\therefore n=9$ is the correct answer.