1965 AHSME Problems/Problem 26
Problem
For the numbers define to be the arithmetic mean of all five numbers; to be the arithmetic mean of and ; to be the arithmetic mean of , and ; and to be the arithmetic mean of and . Then, no matter how , and are chosen, we shall always have:
Solution
We shall begin by eliminating some options through counterexamples. If , then , so , and . Answers (C) and (D) do not allow for , so they can be eliminated. If we set and , then , , , and . Here, , so we can throw out options (A) and (B) as well. Now, we are left with only option .
See Also
1965 AHSC (Problems • Answer Key • Resources) | ||
Preceded by Problem 25 |
Followed by Problem 27 | |
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