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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:
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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: (
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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
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