Question

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

Answer 1

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),

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

$\Rightarrow (2n-9){180}^0=(5n+{235}^0)n$

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

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

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

$\Rightarrow n^2-25n+199=0\Rightarrow n^2-16n-9n+199=0$

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

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

$\therefore n=9|n=16$

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 ($9$)