Question

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

• A. 7
• B. 9
• C. 11
• D. 16

Answer 1

Answer: (B)

Let there be $n$-sides of the polygon, then the sum of its interior angles is given by
$S_n=(2n-4)$ right angle
$=(n-2)\times 180^{\circ }\dots (i)$
Since, the interior angles form an $AP$ with first term $a=120^{\circ }$ and common difference $d=5^{\circ }$
$\therefore S_n=\frac{n}{2}\left[2\times 120^{\circ }+\left(n-1\right)5^{\circ }\right]\dots \left(ii\right)$
From Eqs. $\left(i\right)$ and $\left(ii\right)$,
$\left(n-2\right)\times 180^{\circ }=\frac{n}{2}\left[2\times 120^{\circ }+\left(n-1\right)\times 5^{\circ }\right]$
$\Rightarrow \left(n-2\right)\times 360=n\left(5n+235\right)$
$\Rightarrow n^2-25n+144=0$
$\Rightarrow \left(n-16\right)\left(n-9\right)=0$
$\Rightarrow n=16$ or $n=9$
But, when $n=16$ the last angle
$a_n=a+\left(n-1\right)d$
$=120^{\circ }+\left(16-1\right)\times 5$
$=195^{\circ }$
which is not possible
Hence, $n=9$