A) 8
B) 10
C) 9
D) 6
Correct Answer: C
Solution :
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 \[\frac{n}{2}[2\times 120+(n-1)5]=(n-2)180\] Þ \[{{n}^{2}}-25n+144=0\]Þ\[(n-9)(n-16)=0\]Þ\[n=9,\ 16\] 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\[n=9\].
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