By conditions of the problem, we know that:
y = 15(z-x)/S
Since x< y< z are in Harmonic Sequence, it follows that:
y = 2zx/(z+x)
Accordingly,
15(z-x)/S = 2zx/(z+x)
Or, 15*z^2 - 2Szx - 15*x^2 = 0
Or, z = (2xS +/- 2x*sqrt(S^2+ 225))/30
Or, z/x = (S +/- R)/15; where R= sqrt (S^2 + 225)
If z/x = (S - R)/3, then R-S is negative as S < R, yielding negative z for positive x, which is a contradiction.
Consequently,
z/x = (S + R )/3....(i)
Since RHS of (i) is a rational number, therefore LHS must be rational.
Since S is a positive integer,it follows that LHS is rational only if S^2 + 225 is a perfect square.
Also, since R + S must be greater than R-S, it follows that R - S must correspond to a factor of 225 less than sqrt(225) = 15
Hence, (R+S, R-S) = (225,1); (75,3); (45,5); (25,9), giving:
(R, S) =(113, 112); (39, 36); (25, 20); (17, 8)
Consequently the only possible values of S are 112, 36, 20 and 8.
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VERIFICATION::
As a check, we observe that:
S=8 gives R = 17, so that
z = (8+17)/15 * x = (5/3)*x and y = 2xz/(x+z) = (5/4)* x
Or, x:y:z = 12:15:20, so that:
(x, y, z) = (12p, 15p, 20, where p is a positive integer.
Similarly, S = 20 gives R=15, and:
(x,y,z) = (2p, 3p, 6p)
S = 36 gives R =39, and:
(x,y,z) = (3p, 5p, 15p)
S = 112 gives R = 113 and:
(x,y,z) = (8p, 15p, 120p)
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