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(d) Let $A = \frac{{a + b}}{2},\;G = \sqrt {ab} $ and $H = \frac{{2ab}}{{a + b}}$.

Then ${G^2} = ab$.....(i)

and $AH = \left( {\frac{{a + b}}{2}

Vedclass · Accepted Answer

${G^2} = AH$

Vedclass · Answer

(d) Let $A = \frac{{a + b}}{2},\;G = \sqrt {ab} $ and $H = \frac{{2ab}}{{a + b}}$.

Then ${G^2} = ab$.....(i)

and $AH = \left( {\frac{{a + b}}{2}} \right)\;.\;\frac{{2ab}}{{a + b}} = ab$.....(ii)

From (i) and (ii), we have ${G^2} = AH$.

If the arithmetic, geometric and harmonic means between two positive real numbers be $A,\;G$ and $H$, then

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