1989 AIME Problems/Problem 7
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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 |
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