lf A_1, A₂ ;G₁, G₂and H₁ ,H₂ are two arithmetic, geometric and harmonic means, respectively, between two quantities a and b, then ab is equal to

Since A_1, A₂ are two arithmetic means between a and b, therefore) a, A₁ A₂, b are in A.P. with common difference d given by
$\mathrm{d}=\frac{\mathrm{b}-\mathrm{a}}{2+1}=\frac{\mathrm{b}-\mathrm{a}}{3}\left[\text{ using }\mathrm{d}=\frac{\mathrm{b}-\mathrm{a}}{\mathrm{n}+1}\right]$
Now, 
${\mathrm{A}}_1=\mathrm{a}+\mathrm{d}=\mathrm{a}+\frac{\mathrm{b}-\mathrm{a}}{3}=\frac{2\mathrm{a}+\mathrm{b}}{3}$
and ${\mathrm{A}}_2=\mathrm{a}+2\mathrm{d}=\mathrm{a}+2\left(\frac{\mathrm{b}-\mathrm{a}}{3}\right)=\frac{\mathrm{a}+2\mathrm{b}}{3}$
It is given that G1G₂are two geometric means between a and, b. Therefore) a, G₁ G₂, b are in G.p. with common ratio r given by
$\mathrm{r}={\left(\frac{\mathrm{b}}{\mathrm{a}}\right)}^{\frac{1}{2+1}}={\left(\frac{\mathrm{b}}{\mathrm{a}}\right)}^{1/3}\left[\because \mathrm{r}={\left(\frac{\mathrm{b}}{\mathrm{a}}\right)}^{\frac{1}{\mathrm{n}+1}}\right]$
Now, ${\mathrm{G}}_1=\mathrm{ar}=\mathrm{a}{\left(\frac{\mathrm{b}}{\mathrm{a}}\right)}^{1/3}={\mathrm{a}}^{2/3}{\mathrm{b}}^{1/3}$
and ${\mathrm{G}}_2={\mathrm{ar}}^2=\mathrm{a}{\left(\frac{\mathrm{b}}{\mathrm{a}}\right)}^{2/3}={\mathrm{a}}^{1/3}{\mathrm{b}}^{2/3}$
It is also given that H_1,H₂are two harmonic means between a and b, therefore, a, H₁ H₂, b are in H.p. Hence, 1/a, 1/H1, 1/H₂, 1/b, are in A.P. with common difference D given by
$\mathrm{D}=\frac{\mathrm{a}-\mathrm{b}}{(2+1)\mathrm{ab}}=\frac{\mathrm{a}-\mathrm{b}}{3\mathrm{ab}}\left[\because \mathrm{D}=\frac{\mathrm{a}-\mathrm{b}}{(\mathrm{n}+1)\mathrm{ab}}\right]$
Now, $\frac{1}{{\mathrm{H}}_1}=\frac{1}{\mathrm{a}}+\mathrm{D}=\frac{1}{\mathrm{a}}+\frac{\mathrm{a}-\mathrm{b}}{3\mathrm{ab}}=\frac{\mathrm{a}+2\mathrm{b}}{3\mathrm{ab}}$
or ${\mathrm{H}}_1=\frac{3\mathrm{ab}}{\mathrm{a}+2\mathrm{b}}$
$\frac{1}{{\mathrm{H}}_2}=\frac{1}{\mathrm{a}}+2\mathrm{D}=\frac{1}{\mathrm{a}}+\frac{2(\mathrm{a}-\mathrm{b})}{3\mathrm{ab}}=\frac{2\mathrm{a}+\mathrm{b}}{3\mathrm{ab}}$
We have,
$\begin{array}{l}{\mathrm{A}}_1{\mathrm{H}}_2=\frac{2\mathrm{a}+\mathrm{b}}{3}\times \frac{3\mathrm{ab}}{2\mathrm{a}+\mathrm{b}}=\mathrm{ab},\\ {\mathrm{A}}_2{\mathrm{H}}_1=\frac{\mathrm{a}+2\mathrm{b}}{3}\times \frac{3\mathrm{ab}}{\mathrm{a}+2\mathrm{b}}=\mathrm{ab},\\ {\mathrm{G}}_1{\mathrm{G}}_2=\left({\mathrm{a}}^{2/3}{\mathrm{b}}^{1/3}\right)\left({\mathrm{a}}^{1/3}{\mathrm{b}}^{2/3}\right)=\mathrm{ab}\\ \therefore {\mathrm{A}}_1{\mathrm{H}}_2={\mathrm{A}}_2{\mathrm{H}}_1={\mathrm{G}}_1{\mathrm{G}}_2=\mathrm{ab}\end{array}$