1976 AHSME Problems/Problem 14

Problem 14

The measures of the interior angles of a convex polygon are in arithmetic progression. If the smallest angle is $100^\circ$ , and the largest is $140^\circ$ , then the number of sides the polygon has is

$\textbf{(A) }6\qquad \textbf{(B) }8\qquad \textbf{(C) }10\qquad \textbf{(D) }11\qquad \textbf{(E) }12$

Solution

Let $n$ equal the number of sides the polygon has. The sum of all the interior angles of a polygon is: $180(n-2)$ .

The formula for an arithmetic series is $\frac{n(a_1 + a_n)}{2}$ . Set this equal to $180(n-2)$ and solve. In this case, $a_1=100$ and $a_n=140$ .

Our equation becomes $\frac{n(100+140)}{2} = 180(n-2) \Rightarrow 240n = 360(n-2) \Rightarrow 120n = 720$ .

Simplifying, we get $n = \boxed{\textbf{(A) }6}$ ~jiang147369

1976 AHSME (Problems • Answer Key • Resources)
Preceded by Problem 13	Followed by Problem 15
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1976 AHSME Problems/Problem 14

Problem 14

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