**Question:**

We know that the sum of the interior angles of a triangle is 180°. Show that the sum 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.

**Solution:**

Show that: the sum of the interior angles of polygons with 3, 4, 5, 6, …. sides form an arithmetic progression.

To Find: The sum of the interior angles for a 21 - sided polygon.

Given: That the sum of the interior angles of a triangle is 180°.

NOTE: We know that sum of interior angles of a polygon of side $\mathrm{n}$ is $(\mathrm{n}-2) \times 180^{\circ}$.

Let $a_{n}=(n-2) \times 180^{\circ} \Rightarrow$ Since $a_{n}$ is linear in $n .$ So it forms AP with $3,4,5,6, \ldots \ldots$ sides

$\left\{a_{n}\right.$ is the sum of interior angles of a polygon of side $\left.n\right\}$

By using the above formula, we have

$a_{21}=(21-2) \times 180^{\circ}$

$a_{21}=3420^{\circ}$

So, the Sum of the interior angles for a 21 - sided polygon is equal to 3420°.