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 $d = 5^{\circ}$
$\therefore S_{n}=\frac{n}{2}\left[2\times120^{\circ}+\left(n-1\right)5^{\circ}\right] \ldots\left(ii\right)$
From Eqs. $\left(i\right)$ and $\left(ii\right)$,
$\left(n-2\right)\times180^{\circ}=\frac{n}{2}\left[2\times120^{\circ}+\left(n-1\right)\times5^{\circ}\right]$
$\Rightarrow \left(n-2\right)\times360=n\left(5n+235\right)$
$\Rightarrow n^{2}-25n+144=0$
$\Rightarrow \left(n-16\right)\left(n-9\right)=0$
$\Rightarrow n=16$ or $n=9$
But, when $n = 16$ the last angle
$a_{n}=a+\left(n-1\right)d $
$=120^{\circ}+\left(16-1\right)\times5$
$=195^{\circ}$
which is not possible
Hence, $n=9$