All about
flooble

fun stuff

Get a free chatterbox

Free JavaScript

Avatars
perplexus
dot
info
Home
>
Just Math
More Harmonic Integers (
Posted on 20070312
)
Determine three positive integers x, y and z in
harmonic sequence
such that x<y<z, xy=z, and (x, y, z72) are in arithmetic sequence.
Submitted by
K Sengupta
Rating:
2.0000
(1 votes)
Solution:
(
Hide
)
Since x, y and z are in harmonic sequence with x < y< z, it follows that y = 2xz/(x+z).
Since, xy=z and z must be positive, we obtain: 2*x^2 = x+ z,so that:
z = 2*x^2  x and y = 2x1.
Now, x+ z72 = 2y
Or, 2*x^2  4x  70 = 0.
The only positive root of the above equation is given by x = 7, so that y = 2*71 = 13 and z = 7*13 = 91
Consequently, (x,y,z) =(7,13,91) is the only possible solution to the given problem.
Comments: (
You must be logged in to post comments.
)
Subject
Author
Date
Solution (and a half)
Brian Smith
20070314 20:55:50
Solution
Dej Mar
20070312 11:58:56
spoiler
Ady TZIDON
20070312 08:26:48
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 (5)
Unsolved Problems
Top Rated Problems
This month's top
Most Commented On
Chatterbox:
blackjack
flooble's webmaster puzzle
Copyright © 2002  2018 by
Animus Pactum Consulting
. All rights reserved.
Privacy Information