help geometry

The interior angles of a convex polygon are in an arithmetic progression.
If the smallest angle is 100 degrees and common difference is  10 degrees ,
then find the number of sides.

$\text{Let the sides of the polygon $n$ }$

Arithmetic progression:$\begin{array}{|rcll|} \hline \text{sum} &=& 100^\circ+110^\circ+120^\circ+130^\circ+\ldots+\Big( 100^\circ+(n-1)10^\circ\Big) \\ \text{sum} &=& \left(\dfrac{100^\circ + \Big( 100^\circ+(n-1)10^\circ\Big)}{2}\right) * n \\\\ \text{sum} &=& \left( \dfrac{100^\circ+100^\circ +10^\circ n -10^\circ}{2} \right) * n \\\\ \text{sum} &=& \left( \dfrac{190^\circ +10^\circ n}{2} \right) * n \\\\ \mathbf{\text{sum}} &=& \mathbf{(95^\circ +5^\circ n)*n} \\ \hline \end{array}$

Interior angles of a convex polygon: $\text{sum} = (n-2)*180^\circ$

$\begin{array}{|rcll|} \hline (95^\circ +5^\circ n)*n &=& (n-2)*180^\circ \\ 95 n +5 n^2 &=& 180n -2*180\\ \ldots \\ 5n^2-85n + 360&=& 0 \quad | \quad : 5 \\ \mathbf{n^2-17n + 72} &=& \mathbf{0} \\ \hline n &=& \dfrac{17 \pm \sqrt{17^2- 4 *72 } }{2} \\\\ n &=& \dfrac{17 \pm \sqrt{289-288 } }{2} \\\\ n &=& \dfrac{17 \pm 1 }{2} \\ \mathbf{n}=\mathbf{9} &\text{or}&\mathbf{n}=\mathbf{8} \\ \hline \end{array}$

the EXTERIOR angles of the polygon sum to 360 degrees

   the first exterior angle would be    180 - 100     then 180 - 110    .......  until the sum is 360

180 -100    +    180 -110    +   180 -120   +   180 -130    +   180 -140   +  180 -150    +   180 -160   +   180 - 170 = 360

8 sides      (  to have 9 sides it would require a 180 degree 'angle' )