Question

Show that the sum of the interior angles of polygons with $3,4,5,\mathrm{6......}$ sides from an arithmetic progression and find the sum of the interior angles for a $21$ sided polygon .

• A. $1240$
• B. $1250$
• C. $3340$
• D. $3420$

Answer 1

Answer: (D)

$3420$

We know that sum of interior angles of a polygon of side n is (n-2) x 180 so let tn =(n-2) x 180

$\because tnislinearinn,soitis{n}^{th}termofsomeAP$

$t3=180,n=3$

$t4=2\times 180,n=4$

$t5=3\times 180,n=5$

$t4-t3=180,$

$t5-t4=180$

$commondiffernceis180$

Sum of interior angles of polygon with 3, 4, 5, .sides forms an AP.

Now sum of interior angles of 21 sides polygon is

$t21=\left(21-2\right)\times 180$

$19\times 180$

$3420$