Question

We know that the sum of the interior angles of a triangle is ${180}^{\circ }$. 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.

Answer 1

We know that,

The sum of the interior angles of a polygon with 3 sides, $a_1={180}^{\circ }$

The sum of the interior angles of a polygon with 4 sides, $a_2={360}^{\circ }$

The sum of the interior angles of a polygon with 5 sides, $a_3={540}^{\circ }$

As, $a_2-a_1={360}^{\circ }-{180}^{\circ }={180}^{\circ }$ and $a_3-a_2={540}^{\circ }-{360}^{\circ }={180}^{\circ }$

i.e. $a_2-a_1=a_3-a_2$

So, $a_1,a_2,a_3,\dots \dots$ are in A.P.

Also, $a={180}^{\circ }$ and $d={180}^{\circ }$

Since, the sum of the interior angles of a 3 sides polygon = a

So, the sum of the interior angles of a 21 sides polygon $=a_{19}$

Now,

$a_{19}=a+(19-1)d$

$={180}^{\circ }+18\times {180}^{\circ }$

$={180}^{\circ }+{3240}^{\circ }={3420}^{\circ }$

So, the sum of the interior angles for a 21 sided polygon is ${3420}^{\circ }$