Question

The interior angles of a polygon are in AP. If the smallest angle is 120${}^{\circ }$ and the common difference is 5${}^{\circ }$, then find the possible number of sides of that polygon?

• A. 16
• B. 9
• C. Both a and b
• D. None of these

Answer 1

The correct option is B 9

Sum of the interior angles of a polygon = (n-2) 180

$\frac{n}{2}\times$ (2a+ (n-1)d)= (n-2)180

$\frac{n}{2}\times$[2$\times$(120) + (n-1)5]=(n-2)180

n[48+(n-1)] = (n-2)72

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

n=9 and n=16

When n= 16, the greatest angle will be equal to a+15d = 120 + 15 x 5 = 195 and no interior angle of a polygon can be equal to or greater than 180. Hence answer =9.