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(A) The angles of the polygon form an $A.P.$ with the first term $a = 120^{\circ}$ and common difference $d = 5^{\circ}$.The sum of all interior angle

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$9$

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(A) The angles of the polygon form an $A.P.$ with the first term $a = 120^{\circ}$ and common difference $d = 5^{\circ}$.The sum of all interior angles of a polygon with $n$ sides is given by $S_{n} = 180^{\circ}(n-2)$.Using the sum formula for an $A.P.$,$S_{n} = \frac{n}{2}[2a + (n-1)d]$.Equating the two,we get $\frac{n}{2}[2(120^{\circ}) + (n-1)5^{\circ}] = 180^{\circ}(n-2)$.Multiplying by $2$,$n[240 + 5n - 5] = 360(n-2)$.$n(5n + 235) = 360n - 720$.$5n^{2} + 235n = 360n - 720$.$5n^{2} - 125n + 720 = 0$.Dividing by $5$,$n^{2} - 25n + 144 = 0$.Factoring the quadratic equation,$(n-9)(n-16) = 0$.Thus,$n = 9$ or $n = 16$.However,for $n = 16$,the largest angle is $a + (n-1)d = 120^{\circ} + 15(5^{\circ}) = 120^{\circ} + 75^{\circ} = 195^{\circ}$. Since an interior angle of a convex polygon must be less than $180^{\circ}$,$n = 16$ is rejected.Therefore,the number of sides is $n = 9$.

The difference between any two consecutive interior angles of a polygon is $5^{\circ}$. If the smallest angle is $120^{\circ}$,find the number of the sides of the polygon.

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