1968 AHSME Problems/Problem 20

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Problem

The measures of the interior angles of a convex polygon of $n$ sides are in arithmetic progression. If the common difference is $5^{\circ}$ and the largest angle is $160^{\circ}$, then $n$ equals:

$\text{(A) } 9\quad \text{(B) } 10\quad \text{(C) } 12\quad \text{(D) } 16\quad \text{(E) } 32$

Solution

The formula for the sum of the angles in any polygon is $180(n-2)$. Because this particular polygon is convex and has its angles in an arithmetic sequence with its largest angle being $160$, we can find the sum of the angles.

$a_{n}=160$

$a_{1}=160-5(n-1)$

Plugging this into the formula for finding the sum of an arithmetic sequence...

$n(\frac{160+160-5(n-1)}{2})=180(n-2)$.

Simplifying, we get $n^2+7n-144$.

Since we want the positive solution to the quadratic, we can easily factor and find the answer is $n=\boxed{9}$.

Hence the answer is$\fbox{A}$

See also

1968 AHSME (ProblemsAnswer KeyResources)
Preceded by
Problem 19
Followed by
Problem 21
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