Question

The interior angles of a convex polygon are in AP . The smallest angle is ${120}^{\circ }$ & the common difference is $5^{\circ }$. Find the number of sides of the polygon

• A. 9
• B. 16
• C. 12
• D. none of these

Answer 1

Answer: (A)

9

Let the no. of sides =n

sum of the interior angles of a polygon $=(n-2)\times {180}^0$.......(1)

$\because$ Smallest angle $={120}^0$

$\therefore$ the angles are

${120}^0,{120}^0+5^0,{120}^0+{10}^0$,.......

First term $={120}^0$, common diff= $2\times 5^0$

=${10}^0$

sum of nterm of a A.P.

$=\frac{n}{2}[2a+(n-1\left)d\right]$

$=\frac{n}{2}[2\times {120}^0+(n-1\left)5\right]$

$=\frac{n}{2}[{240}^0+5n-5^0]$

$=\frac{n}{2}[{235}^0+5n]$.......(ii)

since (i)=(ii),

$\therefore (n-2)\times {180}^0=\frac{n}{2}[{235}^0+5n]$

$\Rightarrow (2n-4){180}^0=(5n+{235}^0)n$

$\Rightarrow 360n-{720}^0=5n^2+235n$

$\Rightarrow 5n^2+235n-360n+720=0$

$\Rightarrow n^2-125n+720=0$

$\Rightarrow n^2-25n+144=0\Rightarrow n^2-16n-9n+144=0$

$n(n-16)-9(n-16)=0$

$\Rightarrow (n-9)(n-16)=0$

$\therefore n=9|n=16$

For, $n=16$

We get

$a_{16}=a+(16-1)d$

$=120+15\times 5$

$=120+75$

$={195}^0>{180}^0$

which is not possible

So, no of sides are (9)