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 --> 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.