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... | ||
− | <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>. | ||
Line 28: | Line 28: | ||
== See also == | == See also == | ||
− | {{AHSME box|year=1968|num-b=19|num-a=21}} | + | {{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 20:07, 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 |
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