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Progressively Geometric (Posted on 2009-09-19) Difficulty: 2 of 5
Prove that 11, 12 and 13 can never be three terms (not necessarily consecutive) of a geometric progression, irrespective of whether the common ratio is real or complex.

No Solution Yet Submitted by K Sengupta    
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Solution Corrected Proof (spoiler) | Comment 3 of 4 |
Assume 11 = ar^x
            12 = ar^y
            13 = ar^z  for x, y, z integers
       
Then
  12/11=r^(y-x), so ln(12/11)=(y-x)*ln(r)
  Similarly,              ln(13/12)=(z-y)*ln(r)
  
  Therefore,
  ln(12/11)/ln(13/12) = (y-x)/(z-y)
  
But the left hand side is irrational, and the right hand side is rational, so we have a contradiction.  Therefore, 11, 12 and 13 cannot be part of the same geometric progression.

  Posted by Steve Herman on 2009-09-20 02:24:39
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