2009 IMO Problems/Problem 3
Problem
Suppose that is a strictly increasing sequence of positive integers such that the subsequences
are both arithmetic progressions. Prove that the sequence is itself an arithmetic progression.
Author: Gabriel Carroll, USA
Solution
then
S(s2) - S(s1) = S(s3) - S(s2) 2S(s2) = S(s1) + S(s3)
i.s.w.
2S(s2+1) = S(s1+1) + S(s3+1) 2S(s2) = S(s1) + S(s3)
put S(3) = b, S(2) = a, S(1) = k
--> 2S(k+1) = S(a+1) + S(b+1) 2S(k) = S(a) + S(b) **if there is S(x) = a,b,k (which is an arithmetic sequence) --> we have 2S(k) = S(a) + S(b) ... (1), 2S(k+1) = S(a+1) + S(b+1) ...(2) but also, by (2), we have 2S(k+2) = S(a+2) + S(b+2) .... we get 2S(b+a) = S(z) + S(y)
therefore, every S(n) is an arithmetic sequence. Q.E.D.
See Also
2009 IMO (Problems) • Resources | ||
Preceded by Problem 2 |
1 • 2 • 3 • 4 • 5 • 6 | Followed by Problem 4 |
All IMO Problems and Solutions |