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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 A little different solution Comment 4 of 4 |
Suppose these can be terms in a GP with common ratio r

r^a = 12/11

r^b = 13/12

where a, b are integers

=> (r^a)^b  - (r^b)^a = 0
=>(12/11)^b - (13/12)^a = 0
=> 12^(a+b) - 13^a * 11^b = 0
=> 3^(a+b) * 2^2(a+b) = 13^a * 11^b

As bases are not equal, LHS RHS can't be equal. This is a contradiction.

So, 11, 12 and 13 can never be three terms of a GP

  Posted by Praneeth on 2009-09-20 06:27:34
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