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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.
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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°,
\[\text{ As, } a_2 - a_1 = 360^\circ - 180^\circ = 180^\circ \text { and } a_3 - a_2 = 540^\circ - 360^\circ= 180^\circ\]
\[\text { i . e } . a_2 - a_1 = a_3 - a_2 \]
\[\text { So }, a_1 , a_2 , a_3 , . . . \text { are in A . P } . \]
\[\text { Also, } a = 180^\circ \text { and }d = 180^\circ\]
\[\text { Since, the sum of the interior angles of a 3 sided polygon } = a\]
\[\text { So, the sum of the interior angles of a 21 sided polygon }= a_{19} \]
\[\text { Now, } \]
\[ a_{19} = a + \left( 19 - 1 \right)d\]
\[ = 180^\circ + 18 \times 180^\circ\]
\[ = 180^\circ + 3240^\circ \]
\[ = 3420^\circ\]
So, the sum of the interior angles for a 21 sided polygon is 3420°.
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