The interior angles of a polygon are in AP. If the smallest angle is 120 ∘ and the common difference is 5 ∘ , then find the possible number of sides of that polygon? Options: A . &nbsp 16 B . &nbsp 9 C . &nbsp Both a and b D . &nbsp None of these

The interior angles of a polygon are in AP. If the smallest angle is 120ȡ

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?

Options:

A . 16

B . 9

C . Both a and b

D . None of these

Answer: Option B
:
B

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

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

$\frac{n}{2}$ * [2*(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 =b

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