# Difference between revisions of "1962 AHSME Problems/Problem 20"

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− | {{ | + | If the angles are in an arithmetic progression, they can be expressed as |

+ | <math>a</math>, <math>a+n</math>, <math>a+2n</math>, <math>a+3n</math>, and <math>a+4n</math> for some real numbers <math>a</math> and <math>n</math>. | ||

+ | Now we know that the sum of the degree measures of the angles of a pentagon is <math>180(5-2)=540</math>. | ||

+ | Adding our expressions for the five angles together, we get <math>5a+10n=540</math>. | ||

+ | We now divide by 5 to get <math>a+2n=108</math>. It so happens that <math>a+2n</math> is one of the angles we defined earlier, so that angle must have a measure of <math>\boxed{108\textbf{ (A)}}</math>. | ||

+ | (In fact, for any arithmetic progression with an odd number of terms, | ||

+ | the middle term is equal to the average of all the terms.) |

## Revision as of 15:11, 16 April 2014

## Problem

The angles of a pentagon are in arithmetic progression. One of the angles in degrees, must be:

## Solution

If the angles are in an arithmetic progression, they can be expressed as , , , , and for some real numbers and . Now we know that the sum of the degree measures of the angles of a pentagon is . Adding our expressions for the five angles together, we get . We now divide by 5 to get . It so happens that is one of the angles we defined earlier, so that angle must have a measure of . (In fact, for any arithmetic progression with an odd number of terms, the middle term is equal to the average of all the terms.)