2005 AMC 12A Problems/Problem 13
Contents
Problem
In the five-sided star shown, the letters , , , and are replaced by the numbers 3, 5, 6, 7 and 9, although not necessarily in that order. The sums of the numbers at the ends of the line segments , , , , and form an arithmetic sequence, although not necessarily in that order. What is the middle term of the arithmetic sequence?
Solution
(i.e., each number is counted twice). The sum will always be , so the arithmetic sequence has a sum of . The middle term must be the average of the five numbers, which is .
Solution 2
Let the terms in the arithmetic sequence be , , , , and . We seek the middle term .
These five terms are , , , , and , in some order. The numbers , , , , and are equal to 3, 5, 6, 7, and 9, in some order, so Hence, the sum of the five terms is But adding all five numbers, we also get , so Dividing both sides by 5, we get , which is the middle term. The answer is (D).
See also
2005 AMC 12A (Problems • Answer Key • Resources) | |
Preceded by Problem 12 |
Followed by Problem 14 |
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All AMC 12 Problems and Solutions |
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