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 20090920 06:27:34 