1989 AIME Problems/Problem 7
Contents
[hide]Problem
If the integer is added to each of the numbers , , and , one obtains the squares of three consecutive terms of an arithmetic series. Find .
Solution 1
Call the terms of the arithmetic progression , making their squares .
We know that and , and subtracting these two we get (1). Similarly, using and , subtraction yields (2).
Subtracting the first equation from the second, we get , so . Substituting backwards yields that and .
Solution 2 (Straighforward, but has big numbers)
Since terms in an arithmetic progression have constant differences,
Solution 3
Let the arithmetic sequence be , , and . Then , but using the difference of squares, . Also, , and using the difference of squares we get . Subtracting both equations gives , , and . Since , and .
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Video Solution by OmegaLearn
https://youtu.be/qL0OOYZiaqA?t=251
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See also
1989 AIME (Problems • Answer Key • Resources) | ||
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Followed by Problem 8 | |
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