We know that the sum of the interior angles of a triangle is`180^0` . Show that the sums of eth interior angles of polygons with `3,4,5,6,` sides for an arithmetic progression. Find the sum of the interior angles of or a 21 sided polygon.

Explanation

We know that sum of interior angles of a polygon of side `n=(2n-4)xx90^(@)=(n-2)xx180^(2)`
Sum of interior angles a polygon with sides 3 is 180.
Sum of interior angles of polygon with side `4=(4-2)xx180^(@)=360^(2)`
Similarly , sum of interior angles of polygon with side 5,6,7,. are `540^(@) ,720 ^(@),900^(@)`.......
the series will be `180^(@),360^(@),540^(@),720^(@),900^(@),......`
Here `a=180^(@)`
` and d=360^(2)-180^(@)=180^(@)`
since common difference is same between two consecutive terms of the series .
so ir from an AP .
we have to find the of interior angles of a 21 sides polygen .
it means , we have to find the 19 th term of the above series .
`a_(19)=a+(19-1)d`
`=180+18xx180=3420`