The interior angles of a polygon are in A.P. | vedclass

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Question

(C) Let the number of sides of the polygon be $n$.The sum of the interior angles of a polygon with $n$ sides is given by $(n - 2) 	imes 180^{\circ}$.

Vedclass · Accepted Answer

$9$

Vedclass · Answer

(C) Let the number of sides of the polygon be $n$.The sum of the interior angles of a polygon with $n$ sides is given by $(n - 2) 	imes 180^{\circ}$.Since the angles are in $A.P.$ with first term $a = 120^{\circ}$ and common difference $d = 5^{\circ}$,the sum of the $n$ angles is $\frac{n}{2} [2a + (n - 1)d]$.Equating the two expressions:$\frac{n}{2} [2(120^{\circ}) + (n - 1)5^{\circ}] = (n - 2)180^{\circ}$$n[240 + 5n - 5] = 360(n - 2)$$5n^2 + 235n = 360n - 720$$5n^2 - 125n + 720 = 0$Dividing by $5$:$n^2 - 25n + 144 = 0$$(n - 9)(n - 16) = 0$So,$n = 9$ or $n = 16$.If $n = 16$,the largest angle is $T_{16} = a + 15d = 120^{\circ} + 15(5^{\circ}) = 120^{\circ} + 75^{\circ} = 195^{\circ}$.Since an interior angle of a convex polygon cannot exceed $180^{\circ}$,$n = 16$ is rejected.Therefore,the number of sides is $n = 9$.

The interior angles of a polygon are in $A.P.$ If the smallest angle is $120^{\circ}$ and the common difference is $5^{\circ}$,then the number of sides is

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