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Question

The interior angles of a polygon are in A.P. If the smallest angle is $$100^o$$ and the common difference is $$4^o$$, find the number of sides ?


Solution

$$\boxed{sum\,of\,interior\,angles\,of\,an\,n\,sided\,polygon=180(n-2)}$$
Angles are in $$AP$$

Smallest angle $$=a=100^0$$

Common difference $$=d=4^0$$

Sum of $$n$$ terms $$\dfrac{n}{2}[2a+(n-1)d]$$

Sum of $$n$$ terms $$\dfrac{n}{2}[200+(n-1)4]$$

$$\boxed{sum\,of\,terms\Rightarrow n[100+2(n-1)]}$$......(2)

$$(1)=(2)$$

$$\Rightarrow n[100+2(n-1)]=180(n-2)$$

$$\Rightarrow 100n+2n^2-2n=180n-360$$

$$\Rightarrow 2n^2=82n-360$$

$$\Rightarrow n^2-41n+180=0$$

$$\Rightarrow n^2-36n-5n+180=0$$

$$\Rightarrow n(n-36)-6(n-36)=0$$

$$\Rightarrow (n-5)(n-36)=0$$

$$\Rightarrow n=5$$ or $$36$$

Mathematics

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