# 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

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

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