Question

Let the difference between any two consecutive interior angles of a polygon is $5^{\circ}$. If the smallest angle is $120^{\circ}$, then the number of sides of polygon is

• A. 9
• B. 8
• C. 7
• D. 16

Answer 1

Answer: (A)

We know that, $S_n=\frac{n}{2}[2 a+(n-1) d]$
Also, we know the sum of interior angles of a polygon having $n$ sides, $S_n=(n-2) 180^{\circ}$
Therefore, $(n-2) 180=\frac{n}{2}[2 \times 120+(n-1)(5)]$
$ (\because a=120, d=5)$
$\Rightarrow (n-2) 180 \times 2 =n(240+5 n-5) $
$\Rightarrow (n-2) 360 =n(5 n+235) $
$\Rightarrow (n-2) 72 =n(n+47)$
$\Rightarrow 72 n-144 =n^2+47 n $
$\Rightarrow n^2+47 n-72 n+144 =0 $
$\Rightarrow n^2-25 n+144 =0$
Now, factorising it by splitting the middle term, we get
$\Rightarrow n^2-(16+9) n+144 =0 $
$\Rightarrow n^2-16 n-9 n+144 =0$
$\Rightarrow n(n-16)-9(n-16) =0 $
$\Rightarrow (n-16)(n-9) =0 $
$\Rightarrow n =9,16$
Only $n=9$ is the required number of sides.
Note If $n=16$,
$\Rightarrow T_{16} =120+(16-1) 5$
$ =120+15 \times 5=120+75=195>180^{\circ}$
which is not possible.
Thus, number of sides in the polygon is 9.