If x 1,;y 1,z 1 are in G.P., t | vedclass

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(b) $x,\;y,\;z$ are in $G.P. $Hence ${y^2} = xz$

$	herefore $$2\log y = \log x + \log z$

$ \Rightarrow $$2(\log y + 1) = (1 + \log x) + (1 + \l

Vedclass · Accepted Answer

$H.P.$

Vedclass · Answer

(b) $x,\;y,\;z$ are in $G.P. $Hence ${y^2} = xz$

$	herefore $$2\log y = \log x + \log z$

$ \Rightarrow $$2(\log y + 1) = (1 + \log x) + (1 + \log z)$

$ \Rightarrow $$1 + \log x,\;1 + \log y,\;1 + \log z$ are in $A.P.$

$ \Rightarrow $$\frac{1}{{1 + \log x}},\;\frac{1}{{1 + \log y}},\;\frac{1}{{1 + \log z}}$ are is $H.P.$

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

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