Difference between revisions of "2004 AIME II Problems/Problem 9"

m
m
Line 4: Line 4:
 
== Solution ==
 
== Solution ==
 
{{solution}}
 
{{solution}}
 +
 
== See also ==
 
== See also ==
* [[2004 AIME II Problems/Problem 8 | Previous problem]]
+
{{AIME box|year=2004|n=II|num-b=8|num-a=10}}
* [[2004 AIME II Problems/Problem 10 | Next problem]]
 
* [[2004 AIME II Problems]]
 

Revision as of 12:27, 19 April 2008

Problem

A sequence of positive integers with $a_1=1$ and $a_9+a_{10}=646$ 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 $n\ge1,$ the terms $a_{2n-1}, a_{2n}, a_{2n+1}$ are in geometric progression, and the terms $a_{2n}, a_{2n+1},$ and $a_{2n+2}$ are in arithmetic progression. Let $a_n$ be the greatest term in this sequence that is less than 1000. Find $n+a_n.$

Solution

This problem needs a solution. If you have a solution for it, please help us out by adding it.

See also

2004 AIME II (ProblemsAnswer KeyResources)
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