2004 AIME II Problems/Problem 9
Problem
A sequence of positive integers with and
is formed so that the first three terms are in geometric progression, the second, third, and fourth terms are in arithmetic progression, and, in general, for all
the terms
are in geometric progression, and the terms
and
are in arithmetic progression. Let
be the greatest term in this sequence that is less than
. Find
Solution
Let ; then solving for the next several terms, we find that
, and in general,
,
, where
.[1]
From , we find that by either the quadratic formula or trial-and-error/modular arithmetic that
. Thus
, and we need to find the largest
such that either
. This happens with
, and this is the
th term of the sequence.
The answer is .
^ We can show this by simultaneous induction: since
and
See also
2004 AIME II (Problems • Answer Key • Resources) | ||
Preceded by Problem 8 |
Followed by Problem 10 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
All AIME Problems and Solutions |
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