1989 AIME Problems/Problem 7
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Contents
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,
Video Solution by OmegaLearn
https://youtu.be/qL0OOYZiaqA?t=251
~ pi_is_3.14
See also
1989 AIME (Problems • Answer Key • Resources) | ||
Preceded by Problem 6 |
Followed by Problem 8 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
All AIME Problems and Solutions |
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