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Home  >>  CBSE XI  >>  Math  >>  Sequences and Series
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The difference between any two consecutive interior angles of a polygon is $5^\circ$. If the smallest angle is $120^ \circ$, then find the number of sides of the polygon.

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  • The sum of the interior angles of $n$ sided polygon = $(n-2).180^{\circ}$
  • The sum of $n$ terms of an A.P=$\large\frac{n}{2}$$[2a+(n-1)d]$
Let the no. of sides of the polygon be $n$
The sum of the interior angles of $n$ sided polygon = $(n-2).180^{\circ}$
It is given that the difference between any two consecutive angles is $5^{\circ}$
$\therefore$ The angles of the polygon are assumed as
$a,a+5,a+10,...........a+(n-1).5$
We know that the the sum of the angles =$ (n-2).180$
$\Rightarrow\:a+(a+5)+(a+10)+...........a+(n-1).5=(n-2).180$
This series is an A.P. with common difference $d=5$
We know that the sum of $n$ terms of an A.P=$\large\frac{n}{2}$$[2a+(n-1)d]$
$\Rightarrow\:\large\frac{n}{2}$$[2a+(n-1)5]=(n-2)180$
Also given that the smallest angle is $120^{\circ}$
$\therefore\:a=120$
$\Rightarrow\:\large\frac{n}{2}$$[2\times 120+(n-1)5]=(n-2)180$
$\Rightarrow\:n[240+5n-5]=(n-2).360$
$\Rightarrow\:5n^2+235n=360n-720$
$\Rightarrow\:5n^2-125n+720=0$
Taking $5$ common
$\Rightarrow\:n^2-25n+144=0$
$\Rightarrow\:n^2-16n-9n+144=0$
$\Rightarrow\:n(n-16)-9(n-16)=0$
$\Rightarrow\:(n-16)(n-9)=0$
$\Rightarrow\:n=16\:\:or\:\:n=9$
answered Mar 5, 2014 by rvidyagovindarajan_1
 

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