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, $S_n=\frac{n}{2}[2a+(n-1)d]$
Also, we know the sum of interior angles of a polygon having $n$ sides, $S_n=(n-2)180^{\circ }$
Therefore, $(n-2)180=\frac{n}{2}[2\times 120+(n-1)(5)]$
$(\because a=120,d=5)$
$\Rightarrow (n-2)180\times 2=n(240+5n-5)$
$\Rightarrow (n-2)360=n(5n+235)$
$\Rightarrow (n-2)72=n(n+47)$
$\Rightarrow 72n-144=n^2+47n$
$\Rightarrow n^2+47n-72n+144=0$
$\Rightarrow n^2-25n+144=0$
Now, factorising it by splitting the middle term, we get
$\Rightarrow n^2-(16+9)n+144=0$
$\Rightarrow n^2-16n-9n+144=0$
$\Rightarrow n(n-16)-9(n-16)=0$
$\Rightarrow (n-16)(n-9)=0$
$\Rightarrow n=9,16$
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.