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 19:00, 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 | |
| 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 • 36 • 37 • 38 • 39 • 40 | ||
| All AHSME Problems and Solutions | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
