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 More Harmonic Integers (Posted on 2007-03-12)
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 (and a half) Comment 3 of 3 |

From the given information in the problem, three equations can be formed:
x*y=z [given]
(1/x)+(1/z)=(2/y) [x,y,z harmonic]
x+z-72=y [x,y,z-72 arithmetic]

Substituting the first equation into the other two yields:
(1/x)+(1/(x*y))=(2/y)
x+x*y-72=y

Solving the latter of these two equations for y yields:
y=(x-72)/(2-x)
Substituting this into the other equation:
(1/x) + (1/(x*((x-72)/(2-x)))) = (2/((x-72)/(2-x)))
(1/x) + (2-x)/(x*(x-72)) = 2*(2-x)/(x-72)
(x-72) + (2-x) = 2x*(2-x)
2x^2 - 4x - 70 = 0

Solving the quadratic gives two solution to the system:
(x,y,z) = (7, 13, 91) or (-5, -11, 55).

The only solution set which fits the constraint x<y<z is (7,13,91).

 Posted by Brian Smith on 2007-03-14 20:55:50
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