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2005 AMC 10A Problems/Problem 17

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Problem

In the five-sided star shown, the letters A, B, C, D, and E are replaced by the numbers 3, 5, 6, 7, and 9, although not necessarily in this order. The sums of the numbers at the ends of the line segments AB, BC, CD, DE, and EA form an arithmetic sequence, although not necessarily in this order. What is the middle term of the sequence?

Image:2005amc10a17.gif

\mathrm{(A) \ } 9\qquad \mathrm{(B) \ } 10\qquad \mathrm{(C) \ } 11\qquad \mathrm{(D) \ } 12\qquad \mathrm{(E) \ } 13

Solution

Since each number is part of 2 numbers in the arithmetic sequence, the sum of the arithmetic sequence is 2(3+5+6+7+9)=60

Since the middle term in an arithmetic sequence is the average of all the terms in the sequence, the middle number is \frac{60}{5}=12\Rightarrow D

See Also

Want to learn how to tackle those tough AMC/AIME/Olympiad algebra problems? Check out Art of Problem Solving's Intermediate Algebra by Richard Rusczyk and Mathew Crawford. Over 1600 problems!
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