2009 IMO Problems/Problem 3

Problem

Suppose that is a strictly increasing sequence of positive integers such that the subsequences

and

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