# 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.