Difference between revisions of "1989 AIME Problems/Problem 7"
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Revision as of 18:16, 4 July 2013
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 | |
| 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
| All AIME Problems and Solutions | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
