Difference between revisions of "1968 AHSME Problems/Problem 20"
m (→See also) |
(→Solution) |
||
| Line 19: | Line 19: | ||
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... | ||
| − | <math>n(\frac{160+160-5(n-1)}{2})=180(n-2)</math>. | + | <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>. | ||
Latest revision as of 19:07, 16 August 2024
Problem
The measures of the interior angles of a convex polygon of 
sides are in arithmetic progression. If the common difference is 
and the largest angle is 
, then equals:
Solution
The formula for the sum of the angles in any polygon is 
. Because this particular polygon is convex and has its angles in an arithmetic sequence with its largest angle being 
, we can find the sum of the angles.
Plugging this into the formula for finding the sum of an arithmetic sequence...
.
Simplifying, we get 
.
Since we want the positive solution to the quadratic, we can easily factor and find the answer is 
.
Hence the answer is
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 | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
