Difference between revisions of "1962 AHSME Problems/Problem 20"
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the middle term is equal to the average of all the terms.) | the middle term is equal to the average of all the terms.) | ||
+ | ==Solution 2== | ||
+ | If we write the five terms as <math>a</math>, <math>a - n</math>, <math>a - 2n</math>, <math>a + n</math> and <math>a + 2n</math>, we can see that adding them up, we get <math>5a = 540</math> through this, we can see that <math>a = 108</math>, <math>\fbox{\textbf{(A)}}</math> | ||
==See Also== | ==See Also== |
Latest revision as of 20:00, 29 October 2020
Contents
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.)
Solution 2
If we write the five terms as , , , and , we can see that adding them up, we get through this, we can see that ,
See Also
1962 AHSC (Problems • Answer Key • Resources) | ||
Preceded by Problem 19 |
Followed by Problem 21 | |
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All AHSME Problems and Solutions |
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