1989 AIME Problems/Problem 7
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
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 .
See also
1989 AIME (Problems • Answer Key • Resources) | ||
Preceded by Problem 6 |
Followed by Problem 8 | |
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