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

We know that 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.                [NCERT EXEMPLAR]

Solution:

We know that,

the sum of the interior angles of a polygon with 3 sides, a1 = 180°,

the sum of the interior angles of a polygon with 4 sides, a2 = 360°,

the sum of the interior angles of a polygon with 5 sides, a3 = 540°,

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, $a_{1}, a_{2}, a_{3}, \ldots$ are in A.P.

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

Since, the sum of the interior angles of a 3 sided polygon $=a$

So, the sum of the interior angles of a 21 sided 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°.

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