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(B) Let $A$,$G$,and $H$ be the arithmetic mean,geometric mean,and harmonic mean of $a$ and $b$ respectively.Given: $A = G + \frac{3}{2}$ and $G = H +

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$135$

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(B) Let $A$,$G$,and $H$ be the arithmetic mean,geometric mean,and harmonic mean of $a$ and $b$ respectively.Given: $A = G + \frac{3}{2}$ and $G = H + \frac{6}{5}$.We know that $G^2 = AH$,so $H = \frac{G^2}{A}$.Substituting $H$ into the second equation: $G = \frac{G^2}{A} + \frac{6}{5} \implies G - \frac{6}{5} = \frac{G^2}{A}$.Since $A = G + \frac{3}{2} = \frac{2G + 3}{2}$,we have $G - \frac{6}{5} = \frac{2G^2}{2G + 3}$.$(5G - 6)(2G + 3) = 10G^2 \implies 10G^2 + 15G - 12G - 18 = 10G^2$.$3G = 18 \implies G = 6$.Then $A = 6 + 1.5 = 7.5 = \frac{15}{2}$ and $H = 6 - 1.2 = 4.8 = \frac{24}{5}$.Since $A = \frac{a+b}{2} = \frac{15}{2}$,we have $a+b = 15$.Since $G = \sqrt{ab} = 6$,we have $ab = 36$.The numbers $a$ and $b$ are roots of $x^2 - 15x + 36 = 0$.$(x - 12)(x - 3) = 0$,so ${a, b} = {12, 3}$.$|a^2 - b^2| = |144 - 9| = 135$.

Consider two positive numbers $a$ and $b$. If the arithmetic mean of $a$ and $b$ exceeds their geometric mean by $\frac{3}{2}$ and the geometric mean of $a$ and $b$ exceeds their harmonic mean by $\frac{6}{5}$,then the absolute value of $(a^2 - b^2)$ is equal to

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