All about flooble | fun stuff | Get a free chatterbox | Free JavaScript | Avatars    
perplexus dot info

Home > Just Math
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    
No Rating

Comments: ( Back to comment list | You must be logged in to post comments.)
possible solution | Comment 1 of 4

not sure if this
we have
r^x=11
r^y=12
r^z=13  for some integers x,y,z
12/11=r^(y-x)
ln(12/11)=(y-x)*ln(r)
from r^z=13 we have z*ln(r)=ln(13) thus
ln(r)=ln(13)/z
thus
ln(12/11)=(y-x)*ln(13)/z
ln(12/11)/ln(13)=(y-x)/z
but the left hand side is simple the log of 12/11 to the base 13, this is an irrational number thus can not be equal to (y-x)/z for integers x,y,z.  Thus 11,12,13 can not be part of the same geometric series.


  Posted by Daniel on 2009-09-19 21:47:49
Please log in:
Login:
Password:
Remember me:
Sign up! | Forgot password


Search:
Search body:
Forums (0)
Newest Problems
Random Problem
FAQ | About This Site
Site Statistics
New Comments (10)
Unsolved Problems
Top Rated Problems
This month's top
Most Commented On

Chatterbox:
Copyright © 2002 - 2017 by Animus Pactum Consulting. All rights reserved. Privacy Information