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Difference between revisions of "1968 AHSME Problems/Problem 20"

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Plugging this into the formula for finding the sum of an arithmetic sequence...
 
Plugging this into the formula for finding the sum of an arithmetic sequence...
  
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<math>n(\frac{160+160-5(n-1)}{2})=180(n-2)</math>.
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<math>n\left(\frac{160+160-5(n-1)}{2}\right)=180(n-2)</math>.
  
 
Simplifying, we get <math>n^2+7n-144</math>.
 
Simplifying, we get <math>n^2+7n-144</math>.
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== See also ==
 
== See also ==
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{{AHSME box|year=1968|num-b=19|num-a=21}}   
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{{AHSME 35p box|year=1968|num-b=19|num-a=21}}   
  
 
[[Category: Introductory Geometry Problems]]
 
[[Category: Introductory Geometry Problems]]
 
{{MAA Notice}}
 
{{MAA Notice}}

Latest revision as of 19:07, 16 August 2024

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...

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

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 AHSC (Problems • Answer Key • Resources)
Preceded by
Problem 19 		Followed by
Problem 21
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 • 26 • 27 • 28 • 29 • 30 • 31 • 32 • 33 • 34 • 35
All AHSME Problems and Solutions

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