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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.

Toolbox:
• 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
$\Rightarrow\:n^2-25n+144=0$
$\Rightarrow\:n^2-16n-9n+144=0$
$\Rightarrow\:n(n-16)-9(n-16)=0$
$\Rightarrow\:(n-16)(n-9)=0$
$\Rightarrow\:n=16\:\:or\:\:n=9$