# JEE Main & Advanced Mathematics Geometric Progression Properties of Arithmetic, Geometric, Harmonic Means Between Two Given Numbers

## Properties of Arithmetic, Geometric, Harmonic Means Between Two Given Numbers

Category : JEE Main & Advanced

Let $A,\,\,G$ and $H$ be arithmetic, geometric and harmonic means of two numbers $a$ and $b$.

Then, $A=\frac{a+b}{2},\,G=\sqrt{ab}$ and $H=\frac{2ab}{a+b}$.

These three means possess the following properties :

(1) $A\ge G\ge H$

$A=\frac{a+b}{2},\,G=\sqrt{ab}$ and $H=\frac{2ab}{a+b}$

$\therefore$ $A-G=\frac{a+b}{2}-\sqrt{ab}=\frac{{{(\sqrt{a}-\sqrt{b})}^{2}}}{2}\ge 0$$\Rightarrow$ $A\ge G$   …..(i)

$G-H=\sqrt{ab}-\frac{2ab}{a+b}=\sqrt{ab}\left( \frac{a+b-2\sqrt{ab}}{a+b} \right)=\frac{\sqrt{ab}}{a+b}{{(\sqrt{a}-\sqrt{b})}^{2}}\ge 0$

$\Rightarrow$ $G\ge H$                                                                     …..(ii)

From (i) and (ii), we get $A\ge G\ge H$.

Note that the equality holds only when $a=b$.

(2) $A,\,\,G,\,\,H$ from a G.P., i.e., ${{G}^{2}}=AH$

$AH=\frac{a+b}{2}\times \frac{2ab}{a+b}=ab={{(\sqrt{ab})}^{2}}={{G}^{2}}$. Hence, ${{G}^{2}}=AH$

(3) The equation having $a$ and $b$ as its roots is

${{x}^{2}}-2Ax+{{G}^{2}}=0$

The equation having $a$ and $b$ its roots is

${{x}^{2}}-(a+b)x+ab=0$

$\Rightarrow$ ${{x}^{2}}-2Ax+{{G}^{2}}=0$,     $\left[ \because A=\frac{a+b}{2}\text{ and }G=\sqrt{ab} \right]$.

The roots $a,\,\,\,b$ are given by $A\pm \sqrt{{{A}^{2}}-{{G}^{2}}}$.

(4) If $A,\,\,G,\,\,H$ re arithmetic, geometric and harmonic means between three given numbers $a,\,\,b$ and $c,$ then the equation having $a,\,\,b,\,\,c$ as its roots is ${{x}^{3}}-3A{{x}^{2}}+\frac{3{{G}^{3}}}{H}x-{{G}^{3}}=0$

where $A=\frac{a+b+c}{3},\,G={{(abc)}^{1/3}}$ and $\frac{1}{H}=\frac{\frac{1}{a}+\frac{1}{b}+\frac{1}{c}}{3}$

$\Rightarrow$ $a+b+c=3A,\,abc={{G}^{3}}$ and $\frac{3{{G}^{3}}}{H}=ab+bc+ca$

The equation having $a,\,\,b,\,\,c$ as its roots is

${{x}^{3}}-(a+b+c){{x}^{2}}+(ab+bc+ca)x-abc=0$

$\Rightarrow$ ${{x}^{3}}-3A{{x}^{2}}+\frac{3{{G}^{3}}}{H}x-{{G}^{3}}=0$.

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