Chapter A.9, Problem 126AYU

To determine

To show: That the harmonic mean of $a$ and $b$ equals the geometric mean squared divided by the arithmetic mean.

Answer to Problem 126AYU

Arithmetic mean is $AM = \frac{a + b}{2}$

Geometric mean is $GM = \sqrt{ab}$

Harmonic mean is $HM = \frac{2ab}{a + b}$

Therefore $\frac{G{M}^2}{AM} = \frac{{\left(\sqrt{ab}\right)}^2}{\frac{a + b}{2}} = ab \times \frac{2}{a + b} = \frac{2ab}{a + b} = HM$

Hence the harmonic mean of $a$ and $b$ equals the geometric mean squared divided by the arithmetic mean.

Explanation of Solution

Arithmetic mean is $AM = \frac{a + b}{2}$

Geometric mean is $GM = \sqrt{ab}$

Harmonic mean is $HM = \frac{2ab}{a + b}$

Therefore $\frac{G{M}^2}{AM} = \frac{{\left(\sqrt{ab}\right)}^2}{\frac{a + b}{2}} = ab \times \frac{2}{a + b} = \frac{2ab}{a + b} = HM$

Hence the harmonic mean of $a$ and $b$ equals the geometric mean squared divided by the arithmetic mean.