Question

The interior angles of a polygon of n sides are in AP. The smallest angle is ${120}^o$ and the common difference is $5^o$. How many possible values can $n$ have?

• A. One
• B. Two
• C. Three
• D. Infinitely many

Answer 1

The correct option is B Two

Sum of interior angles of n-sided polygon $={180}^{\circ }(n-2)$

$\therefore$ According to problem

$\Rightarrow {180}^{\circ }(n-2)={120}^{\circ }+{125}^{\circ }+{130}^{\circ }+....$n terms

$\Rightarrow {180}^{\circ }(n-2)=\frac{n}{2}[2\times {120}^{\circ }+(n-1\left)5^{\circ }\right]$

$\Rightarrow {360}^{\circ }n-{720}^{\circ }={240}^{\circ }n+5^{\circ }{n}^2-5^{\circ }n$

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

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

$\therefore n=16$ or $9$