Question

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

Answer 1

Since the sum of all enternal angles $={360}^o$

we take a decreasing AP, where the first term = ${180}^o$ - original first term and common difference is still $5^o$, but in a decreasing manner, $d=-5^o$

We can apply the AP sum formula $\Rightarrow \frac{n}{2}(2a+(4-1\left)d\right)={360}^o$

$\Rightarrow \frac{n}{2}(120+(n-1\left)\right(-5\left)\right)={360}^o$

$\Rightarrow {720}^o=120n-5n^2+5n$

$\Rightarrow 5n^2-125n+{720}^o=0$

$\Rightarrow n^2-25n+{144}^o=0$

$\Rightarrow n^2-16n-9n+{144}^o=0$

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

$\Rightarrow n=9$ or $n=16$

$\therefore n=9$ as $n=16$ takes each angle (term) decreasing below zero and makes it megative to get to the required sum.