Question

If one A.M., A and two geometric means G₁ and G₂ inserted between any two positive numbers, show that $\frac{G_1^2}{G_2}+\frac{G_2^2}{G_1}=2A$.

Answer 1

$$\begin{gathered} \mathrm{Let}\mathrm{the}\mathrm{two}\mathrm{positive}\mathrm{numbers}\mathrm{be}a\mathrm{and}b. \\ a,A\mathrm{and}b\mathrm{are}\mathrm{in}\mathrm{A}.\mathrm{P}. \\ \therefore 2A=a+b........\left(\mathrm{i}\right) \\ \mathrm{Also},a,G_1,G_2\mathrm{and}b\mathrm{are}\mathrm{in}\mathrm{G}.\mathrm{P}. \\ \therefore r={\left(\frac{b}{a}\right)}^{\frac{1}{3}} \\ \mathrm{Also},G_1=ar\mathrm{and}{G}_2=a{r}^2.........\left(\mathrm{ii}\right) \\ \mathrm{Now},\mathrm{LHS}=\frac{{G_1}^2}{G_2}+\frac{{G_2}^2}{G_1} \\ =\frac{{\left(ar\right)}^2}{a{r}^2}+\frac{{\left(a{r}^2\right)}^2}{ar}\left[\mathrm{Using}\left(\mathrm{ii}\right)\right] \\ =a+a{r}^3 \\ =a+a{\left({\left(\frac{b}{a}\right)}^{\frac{1}{3}}\right)}^3 \\ =a+a\left(\frac{b}{a}\right) \\ =a+b \\ =2A \\ =\mathrm{RHS}\left[\mathrm{Using}\left(\mathrm{i}\right)\right] \end{gathered}$$