The interior angles of a polygon are in AP.

Given:

Interior angles of a polygon are in A.P

Smallest angle $=a=52^{\circ}$

Common difference $=\mathrm{d}=8^{\circ}$

Let the number of sides of a polygon $=n$

Angles are in the following order

$52^{\circ}, 52^{\circ}+d, 52^{\circ}+2 d, \ldots \ldots . ., 52^{\circ}+(n-1) \times d$

Sum of $n$ terms in $A . P=s=\frac{n}{2}\{2 a+(n-1) d\}$

Sum of angles of the given polygon is $\frac{\mathrm{n}}{2}\left\{\left(2 \times 52^{\circ}\right)+(\mathrm{n}-1) \times 8^{\circ}\right\}$.

Hint:

Sum of interior angles of a polygon of $n$ sides is $(n-2) \times 180^{\circ}$

Therefore,

$(\mathrm{n}-2) \times 180^{\circ}=\frac{\mathrm{n}}{2}\left\{104^{\circ}+(\mathrm{n}-1) \times 8^{\circ}\right\}$

$180 n-360=52 n+n(n-1) \times 4$

$4 n^{2}+48 n=180 n-360$

$4 n^{2}-132 n+360=0$

$n^{2}-33 n+90=0$

$(n-3)(n-30)=0$

$n=3 \& n=30$

∴ It can be a triangle or a 30 sided polygon.

The number of sides of the polygon is 3 or 30.