If the aritmetic, geometric and harmonic menas between two positive real numbers be A, G and H, then

$A^{2}$ = GH

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$H^{2}$ = AG

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G = AH

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$G^{2}$ = AH

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Solution

The correct option is D
$G^{2}$ = AH

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

Then $G^{2}$ = ab ..................(i)

and AH = $(\frac{a + b}{2}) . \frac{2 a b}{a + b}$ = ab ................(ii)

From (i) and (ii), we have $G^{2}$ = AH

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If the aritmetic, geometric and harmonic menas between two positive real numbers be A, G and H, then

The correct option is D G2 = AH Let A = a+b2, G =√ab and H = 2aba+b. Then G2 = ab ..................(i) and AH = (a+b2).2aba+b = ab ................(ii) From (i) and (ii), we have G2 = AH

The correct option is D
$G^{2}$ = AH

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

Then $G^{2}$ = ab ..................(i)

and AH = $(\frac{a + b}{2}) . \frac{2 a b}{a + b}$ = ab ................(ii)

From (i) and (ii), we have $G^{2}$ = AH