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If $G$ be the geometric mean between two quantities $A$ and $B$, show that the ratio of the arithmetic and harmonic means of $A$ and $G$ is equal to the ratio of the arithmetic and harmonic means of $G$ and $B$.

$A, G, B$ are in geometric progression. Therefore $\dfrac{B}{G}=\dfrac{G}{A}$, or $G^2=AB$

The arithmetic mean of $A$ and $G$ by definition is $\dfrac{A+G}{2}$, and the harmonic mean $H$ of $A$ and $G$ can be found by recognizing the arithmetical progression is the inverse of the harmonical progression: $A,\dfrac{1}{H}, G$

$\dfrac{1}{H}-A=G-\dfrac{1}{H} \Rightarrow \dfrac{2}{H}=A+G \Rightarrow H=\dfrac{2}{A+G}$

Ratio of arithmetic and harmonic means of $A$ and $G$: $\dfrac{\dfrac{A+G}{2}}{\dfrac{2}{A+G}}=\dfrac{(A+G)^2}{4}$

Arithmetic mean of $G$ and $B$: $\dfrac{G+B}{2}$

Harmonic mean: $G,\dfrac{1}{H}, B \Rightarrow \dfrac{1}{H}-G=B-\dfrac{1}{H} \Rightarrow \dfrac{2}{H}=G+B \Rightarrow H=\dfrac{2}{G+B}$

Ratio fo arithmetic and harmonic mean of $G$ and $B$: $\dfrac{\dfrac{G+B}{2}}{\dfrac{2}{G+B}} \Rightarrow \dfrac{(G+B)^2}{4}$

$\dfrac{(A+G)^2}{4}=\dfrac{(G+B)^2}{4} \Rightarrow A+G=G+B \Rightarrow A=B$

This result is wrong because $A,G,B$ are in geometric progression, so $A \neq B$. I also don't really understand if $A,G,B$ are in geometric progression, how could it also be in arithmetical progression at the same time (if that's what the question means). Thanks.

ronald christenkkson
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2 Answers2

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You have an error in

Ratio of arithmetic and harmonic means of $A$ and $G$: $\dfrac{\dfrac{A+G}{2}}{\dfrac{2}{A+G}}$

as this should instead have been $$\dfrac{\dfrac{A+G}{2}}{\dfrac{2}{\frac1A+\frac1G}}$$ which is $$\dfrac{(A+G)^2}{4AG} =\dfrac{(A+A^{1/2}B^{1/2})^2}{4A^{3/2}B^{1/2}} =\dfrac{(A^{1/2}+B^{1/2})^2}{4A^{1/2}B^{1/2}} =\dfrac{(A^{1/2}B^{1/2}+B)^2}{4A^{1/2}B^{3/2}}=\dfrac{(G+B)^2}{4GB}$$

i.e. $$\dfrac{\dfrac{G+B}{2}}{\dfrac{2}{\frac1G+\frac1B}}$$ the ratio of arithmetic and harmonic means of $G$ and $B$

Henry
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The harmonic mean of $x$ and $y$ is not $\frac2{x+y}$ but $\frac2{\frac1x+\frac1y}.$

What you must prove thus reduces to$$(A+G)\left(\frac1A+\frac1G\right)=(B+G)\left(\frac1B+\frac1G\right)$$ or equivalently: $$2+\frac GA+\frac AG=2+\frac GB+\frac BG.$$

For this, simply recall that $G=\sqrt{AB}$.

Anne Bauval
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