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Consider a set $a$ with $t$ terms. Knowing the arithmetic mean, the geometric mean, and the harmonic mean of $a$, could I somehow solve for $t$? If not, what more information would I need to know?

4yl1n
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For a nontrivial example, let $$(x_1, x_2, x_3) = (2, 1, \tfrac{1}{2}),$$ so that the arithmetic, geometric, and harmonic means are $$(\tfrac{7}{6}, 1, \tfrac{6}{7}).$$ Then consider $$(y_1, y_2) = (y, \tfrac{1}{y})$$ for some $y > 1$. Their arithmetic, geometric, and harmonic means are $$\left(\frac{y+y^{-1}}{2}, 1, \frac{2}{y+y^{-1}}\right),$$ hence $y$ satisfies $$\frac{7}{6} = \frac{y + y^{-1}}{2}$$ or $$y = \frac{7 + \sqrt{13}}{6}.$$ This furnishes a counterexample in which given the three means, there are two distinct sets of different sizes that give the same means.

heropup
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    For an example where the $x_i$ and $y_i$ are rational take $(3/8, 1, 8/3)$ and $(4/9, 9/4)$ where the means are $(97/72, 1, 72/97)$. – Robert Israel Apr 12 '20 at 22:26
  • @heropup So what more would I need to know? – 4yl1n Apr 13 '20 at 20:08
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    @4yl1n Based on the feedback you have received so far, it should become increasingly obvious that it is in general impossible to determine the sample size based on the three means, and that in order to do so, you would need an amount of information that is comparable to the sample size itself. – heropup Apr 13 '20 at 20:12