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]

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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°,
.
.
.

$As, a_{2} - a_{1} = 360 ° - 180 ° = 180 ° and a_{3} - a_{2} = 540 ° - 360 ° = 180 ° i . e . a_{2} - a_{1} = a_{3} - a_{2} So, a_{1}, a_{2}, a_{3}, . . . are in A . P . Also, a = 180 ° and d = 180 ° Since, the sum of the interior angles of a 3 sided polygon = a So, the sum of the interior angles of a 21 sided polygon = a_{19} Now, a_{19} = a + (19 - 1) d = 180 ° + 18 \times 180 ° = 180 ° + 3240 ° = 3420 °$

So, the sum of the interior angles for a 21 sided polygon is 3420°.

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