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.

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