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.

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.