Question

The interior angles of a polygon are in A.P if the smallest angle be ${120}^{\circ }$ and the common difference be 5, then the number of sides is

• A. $8$
• B. $10$
• C. $9$
• D. $6$

Answer 1

Answer: (C)

$9$

Let if the polygon has n side $\therefore$ sum its angles = $(n-2){180}^{\circ }........\left(i\right)$
Smallest angle (a) = ${120}^{\circ }$ Common difference (d) = 5 No.of sides = n
ATP Sum of its angles = $\frac{n}{2}[2\times 120+(n-1\left)5\right]........\left(ii\right)$
By (i) & (ii)
$(n-2){180}^{\circ }=\frac{n}{2}[240+(n-1\left)5\right]$
$360n-720=240n+5n^2-5n$
$5n^2-125n+720=0$
$n^2-25n+144=0$
$(n-16)(n-9)=0$ $n=16$ or $n=9$
But n = 16 is not possible for regular polygon because $a_{16}=a+15d=120+15\times 5={195}^{\circ }$