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 1
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
Solution 2
Let . It is apparent that the sequence grows relatively fast, so we start trying positive integers to see what can be. Finding that works, after bashing out the rest of the terms we find that and , hence our answer is .
Solution 3
We can find the value of by its bounds using three conditions:
- (note that the sequence must be increasing on all terms, not monotonically increasing)
- , so necessarily is a factor of , which factorizes to
Rearranging conditions 1 and 2, we get:
trying all the terms from the third condition, it is clear that is the only solution. Then we can calculate the next few terms from there since we have as well, to find that and , thus we have our answer of .
~KafkaTamura
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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