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# If A is the arithmetic mean and G1,

Question:

If A is the arithmetic mean and G1, G2 be two geometric means between any two numbers, then prove that

$2 \mathrm{~A}=\frac{\mathrm{G}_{1}^{2}}{\mathrm{G}_{2}}+\frac{\mathrm{G}_{2}^{2}}{\mathrm{G}_{1}}$

Solution:

Given $\mathrm{A}$ is the arithmetic mean and $\mathrm{G}_{1}, \mathrm{G}_{2}$ be two geometric means between any two numbers

Let the two numbers be ' $\mathrm{a}$ ' and ' $\mathrm{b}$ '

The arithmetic mean is given by $A=\frac{a+b}{2}$ and the geometric mean is given by $\mathrm{G}=\sqrt{\mathrm{ab}}$

We have to insert two geometric means between $\mathrm{a}$ and $\mathrm{b}$

Now that we have the terms $a, G_{1}, G_{2}, b$

$\mathrm{G}_{1}$ will be the geometric mean of a and $\mathrm{G}_{2}$ and $\mathrm{G}_{2}$ will be the geometric mean of $\mathrm{G}_{1}$ and $\mathrm{b}$

Hence $\mathrm{G}_{1}=\sqrt{\mathrm{aG}_{2}}$ and $\mathrm{G}_{2}=\sqrt{\mathrm{G}_{1} \mathrm{~b}}$

Square $\mathrm{G}_{1}=\sqrt{\mathrm{aG}_{2}}$

$\Rightarrow \mathrm{G}_{1}^{2}=\mathrm{aG}_{2}$

Put $\mathrm{G}_{2}=\sqrt{\mathrm{G}_{1} \mathrm{~b}}$

$\Rightarrow \mathrm{G}_{1}^{2}=\mathrm{a} \sqrt{\mathrm{G}_{1} \mathrm{~b}}$

Squaring on both sides we get

$\Rightarrow \mathrm{G}_{1}^{4}=\mathrm{a}^{2}\left(\mathrm{G}_{1} \mathrm{~b}\right)$

$\Rightarrow \mathrm{G}_{1}^{3}=\mathrm{a}^{2} \mathrm{~b}$

$\Rightarrow \mathrm{G}_{1}=\mathrm{a}^{\frac{2}{3}} \mathrm{~b}^{\frac{1}{3}}$

Put value of $\mathrm{G}_{1}$ in $\mathrm{G}_{2}=\sqrt{\mathrm{G}_{1} \mathrm{~b}}$

$\Rightarrow \mathrm{G}_{2}=\sqrt{\mathrm{a}^{\frac{2}{3}} \mathrm{~b}^{\frac{1}{3}} \mathrm{~b}}$

$=\left(a^{\frac{2}{3}} b^{\frac{1}{3}+1}\right)^{\frac{1}{2}}$

On simplification we get

$=\left(a^{\frac{2}{3}} b^{\frac{4}{3}}\right)^{\frac{1}{2}}$

$=a^{\frac{1}{3}} b^{\frac{2}{3}}$

Now we have to prove that $2 \mathrm{~A}=\frac{\mathrm{G}_{1}^{2}}{\mathrm{G}_{2}}+\frac{\mathrm{G}_{2}^{2}}{\mathrm{G}_{1}}$

Consider RHS

$\Rightarrow \mathrm{RHS}=\frac{\mathrm{G}_{1}^{2}}{\mathrm{G}_{2}}+\frac{\mathrm{G}_{2}^{2}}{\mathrm{G}_{1}}$

Substitute values of $G_{1}$ and $G_{2}$ from 1 and 2

$\Rightarrow \mathrm{RHS}=\frac{\left(\mathrm{a}^{\frac{2}{3}} \mathrm{~b}^{\frac{1}{3}}\right)^{2}}{\mathrm{a}^{\frac{1}{3}} \mathrm{~b}^{\frac{2}{3}}}+\frac{\left(\mathrm{a}^{\frac{1}{3}} \mathrm{~b}^{\frac{2}{3}}\right)^{2}}{\mathrm{a}^{\frac{2}{3}} \mathrm{~b}^{\frac{1}{3}}}$

$=\frac{a^{\frac{4}{3}} b^{\frac{2}{3}}}{a^{\frac{1}{3}} b^{\frac{2}{3}}}+\frac{a^{\frac{2}{3}} b^{\frac{4}{3}}}{a^{\frac{2}{3}} b^{\frac{1}{3}}}$

Taking LCM and simplifying we get

$=a^{\frac{4}{3}-\frac{1}{3}} b^{\frac{2}{3}}-\frac{2}{3}+a^{\frac{2}{3}}-\frac{2}{3} b^{\frac{4}{3}}-\frac{1}{3}$

$\Rightarrow$ RHS $=a+b$

Divide and multiply by 2

$\Rightarrow \mathrm{RHS}=2 \frac{\mathrm{a}+\mathrm{b}}{2}$

But $A=\frac{a+b}{2}$

Therefore

⇒ RHS = 2A

Hence RHS = LHS

Hence proved