We know the sum of the interior angles of a triangle is 180°. Show that the sums of the interior angles of polygons with 3, 4, 5, 6, ... sides form an arithmetic progression. Find the sum of the interior angles for a 21 sided polygon​

Dear student

The sum of interior angles of a polygon is given by

For a polygon of n sides, sum of angles = (n-2)$\mathrm{\pi }$

So, for a polygon of (n+1) sides, sum of angles =(n+1-2)$\mathrm{\pi }$=(n-1)$\mathrm{\pi }$

Difference = (n-1)$\mathrm{\pi }$-(n-2)$\mathrm{\pi }$=$\mathrm{\pi }$ = constant

Hence the sums of the interior angles of polygons with 3, 4, 5, 6, ... sides form an arithmetic progression

For a polygon of 21 sides, sum of angles = (n-2)$\mathrm{\pi }$=sum of angles = (21-2)$\mathrm{\pi }$=19$\mathrm{\pi }$

Regards