Difference between revisions of "2009 IMO Problems/Problem 3"
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2S(k) = S(a) + S(b) | 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) | + | **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) | but also, by (2), we have 2S(k+2) = S(a+2) + S(b+2) | ||
.... | .... | ||
we get 2S(b+a) = S(z) + S(y) | we get 2S(b+a) = S(z) + S(y) | ||
therefore, | 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 | ||
