If 1, log_{y}x, log_{z}y, -15log_{x}z are in | vedclass

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(D) Let $d$ be the common difference of the $A.P.$Then,$\log_{y}x = 1 + d \Rightarrow x = y^{1+d}$$\log_{z}y = 1 + 2d \Rightarrow y = z^{1+2d}$$-15\lo

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(D) Let $d$ be the common difference of the $A.P.$Then,$\log_{y}x = 1 + d \Rightarrow x = y^{1+d}$$\log_{z}y = 1 + 2d \Rightarrow y = z^{1+2d}$$-15\log_{x}z = 1 + 3d$ $\Rightarrow \log_{x}z = -\frac{1+3d}{15}$ $\Rightarrow z = x^{-\frac{1+3d}{15}}$Substituting the values,we get $x = (z^{1+2d})^{1+d} = z^{(1+2d)(1+d)} = (x^{-\frac{1+3d}{15}})^{(1+2d)(1+d)}$This implies $1 = -\frac{(1+d)(1+2d)(1+3d)}{15}$$(1+d)(1+2d)(1+3d) = -15$$(1+3d+2d^2)(1+3d) = -15$$1 + 3d + 3d + 9d^2 + 2d^2 + 6d^3 = -15$$6d^3 + 11d^2 + 6d + 16 = 0$By inspection,$d = -2$ is a root: $6(-8) + 11(4) + 6(-2) + 16 = -48 + 44 - 12 + 16 = 0$For $d = -2$,we have $\log_{y}x = 1 - 2 = -1 \Rightarrow x = y^{-1}$$\log_{z}y = 1 + 2(-2) = -3 \Rightarrow y = z^{-3}$Since $x = y^{-1}$ and $y = z^{-3}$,then $x = (z^{-3})^{-1} = z^{3}$Thus,$x = y^{-1}$,$y = z^{-3}$,and $x = z^{3}$ are all true.

If $1, \log_{y}x, \log_{z}y, -15\log_{x}z$ are in $A.P.$,then

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