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.

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).