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More Harmonic Integers (Posted on 2007-03-12) Difficulty: 2 of 5
Determine three positive integers x, y and z in harmonic sequence such that x<y<z, xy=z, and (x, y, z-72) are in arithmetic sequence.

See The Solution Submitted by K Sengupta    
Rating: 2.0000 (1 votes)

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Solution Solution | Comment 2 of 3 |

Solving, first, for the equation x2 + x(n - 1) - (2n + 72) = 0
I found the positive arithmetic sequence to be
(7, 13, [91-72]) where x=7 and n=6; with the harmonic sequence confirmed with (1/13 - 1/7) =  (1/91 - 1/13).

Therefore, x = 7; y = 13; z = 91

The other solutions of positive integers that solve for the arithmetic sequence where xy=z -- eg., (4, 34, 136) and (3, 69, 207), -- do not produce the harmonic sequence.


  Posted by Dej Mar on 2007-03-12 11:58:56
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