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If the A.M of two numbers exceeds their G.M by 2 and the G.M exceeds their H.M by 8/5, then the numbers are

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Solutions : Let the two numbers be $a$ and $b$ , where $a> b$ .

Let their $A.M$ , $G.M$ and $H.M$ be the respectively $A,G$ and $H.$

Then $A=G+2$ and $G=H+frac85$ or $H=G-frac85.$

$herefore$ $AH=G^2Rightarrow (G+2)(G-frac85)=G^2$

$Rightarrow$ $G=8=sqrtab$

$herefore$ $A=8+2=10=fraca+b2Rightarrow a+b=20$ $.......(1)$

$(a-b)^2=(a+b)^2-4ab=400-256=144$ $.......(2)$

From $(1)$ and $(2)$ We get , $a=16,b=4$

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Deependra Verma

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