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 $n$.

Then the sum of interior angles of the polygon

$ = (2n - 4)\frac{\pi }{2} = (n - 2)\pi $

S

Vedclass · Accepted Answer

$9$

Vedclass · Answer

(c) Let the number of sides of the polygon be $n$.

Then the sum of interior angles of the polygon

$ = (2n - 4)\frac{\pi }{2} = (n - 2)\pi $

Since the angles are in $A.P. $ and $a = {120^o},\;d = 5$,

therefore $\frac{n}{2}[2 	imes 120 + (n - 1)5] = (n - 2)180$

==> ${n^2} - 25n + 144 = 0$

==> $(n - 9)(n - 16) = 0$

==> $n = 9,\;16$

But $n = 16$ gives ${T_{16}} = a + 15d = {120^o} + {15.5^o} = {195^o}$,

which is impossible as interior angle cannot be greater than ${180^o}$.

Hence $n = 9$.

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

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