2009 IMO Problems/Problem 3: Difference between revisions
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''Author: Gabriel Carroll, USA'' | ''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== | |||
{{IMO box|year=2009|num-b=2|num-a=4}} | |||
Latest revision as of 00:16, 19 November 2023
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 | ||
