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

Home > Just Math
AM and GM equality (Posted on 2008-01-23) Difficulty: 2 of 5
If Arithmetic Mean of 1st k terms of Arithmetic Progression with 1st term a and common difference d is equal to the Geometric Mean of 1st k terms of Geometric Progression with 1st term a and common ratio r, then find the range of r for which a≈2d/(r-1).

See The Solution Submitted by Praneeth    
Rating: 5.0000 (1 votes)

Comments: ( Back to comment list | You must be logged in to post comments.)
re: ...Now where do I go??? Comment 3 of 3 |
(In reply to ...Now where do I go??? by Dej Mar)

Dej Mar?
In your  expression:
a = (d*0 + ... + d*(k-1))/[k*(r0 * ... * r(k-1)](1/k) - 1]
   d is in the numerator and r the denominator,
but in your last statement you have reversed them.

  Posted by brianjn on 2008-01-28 21:02:09

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 (3)
Unsolved Problems
Top Rated Problems
This month's top
Most Commented On

Chatterbox:
Copyright © 2002 - 2024 by Animus Pactum Consulting. All rights reserved. Privacy Information