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.
Solving, first, for the equation x^{2} + x(n  1)  (2n + 72) = 0
I found the positive arithmetic sequence to be
(7, 13, [9172]) 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 20070312 11:58:56 