If x 1, y 1, z 1 are in G.P.,then frac{ | vedclass

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(B) Given that $x, y, z$ are in $G.P.$,we have $y^2 = xz$.Taking the natural logarithm on both sides,we get $2 \ln y = \ln x + \ln z$.Adding $2$ to bo

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$H.P.$

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(B) Given that $x, y, z$ are in $G.P.$,we have $y^2 = xz$.Taking the natural logarithm on both sides,we get $2 \ln y = \ln x + \ln z$.Adding $2$ to both sides,we get $2 + 2 \ln y = 2 + \ln x + \ln z$.This can be rewritten as $2(1 + \ln y) = (1 + \ln x) + (1 + \ln z)$.This implies that $(1 + \ln x), (1 + \ln y), (1 + \ln z)$ are in $A.P.$Since the reciprocals of terms in $A.P.$ are in $H.P.$,it follows that $\frac{1}{1 + \ln x}, \frac{1}{1 + \ln y}, \frac{1}{1 + \ln z}$ are in $H.P.$

If $x > 1, y > 1, z > 1$ are in $G.P.$,then $\frac{1}{1 + \ln x}, \frac{1}{1 + \ln y}, \frac{1}{1 + \ln z}$ are in

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