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.
Best Answer
You have an error in
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 $B$