If x 1, y 1, z 1 are in geometric progr | vedclass

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(B) Given that $x, y, z$ are in geometric progression,we have $y^2 = xz$. Taking the natural logarithm on both sides,we get $\ln(y^2) = \ln(xz)$,which

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Harmonic Progression

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(B) Given that $x, y, z$ are in geometric progression,we have $y^2 = xz$. Taking the natural logarithm on both sides,we get $\ln(y^2) = \ln(xz)$,which simplifies to $2 \ln y = \ln x + \ln z$. This implies that $\ln x, \ln y, \ln z$ are in arithmetic progression. Adding $1$ to each term,we get $1 + \ln x, 1 + \ln y, 1 + \ln z$,which are also in arithmetic progression. Since the reciprocals of terms in an arithmetic progression form a harmonic progression,the terms $\frac{1}{1 + \ln x}, \frac{1}{1 + \ln y}, \frac{1}{1 + \ln z}$ are in harmonic progression.

If $x > 1, y > 1, z > 1$ are in geometric progression,then in which progression are $\frac{1}{1 + \ln x}, \frac{1}{1 + \ln y}, \frac{1}{1 + \ln z}$?

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