If 1, log_y x, log_z y, -15 log_x z are in A. | vedclass

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

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All the above

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(D) Let $d$ be the common difference of the $A.P.$Then,$\log_y x = 1 + d \implies x = y^{1+d}$$\log_z y = 1 + 2d \implies y = z^{1+2d}$$-15 \log_x z = 1 + 3d \implies \log_x z = -\frac{1+3d}{15} \implies z = x^{-(1+3d)/15}$Substituting these,we get $x = y^{1+d} = (z^{1+2d})^{1+d} = z^{(1+2d)(1+d)}$.Also,$z = x^{-(1+3d)/15} \implies x = z^{-15/(1+3d)}$.Equating the exponents of $z$: $(1+2d)(1+d) = -\frac{15}{1+3d}$.$(1+d)(1+2d)(1+3d) = -15$.Expanding the product: $(1 + 3d + 2d^2)(1 + 3d) = 1 + 3d + 3d + 9d^2 + 2d^2 + 6d^3 = 6d^3 + 11d^2 + 6d + 1 = -15$.$6d^3 + 11d^2 + 6d + 16 = 0$.Testing $d = -2$: $6(-8) + 11(4) + 6(-2) + 16 = -48 + 44 - 12 + 16 = 0$.Thus,$d = -2$ is a root.For $d = -2$: $\log_y x = 1 - 2 = -1 \implies x = y^{-1}$.$\log_z y = 1 + 2(-2) = -3 \implies y = z^{-3}$.Since $x = y^{-1}$ and $y = z^{-3}$,we have $x = (z^{-3})^{-1} = z^3$.Therefore,all the given options are correct.

If $1, \log_y x, \log_z y, -15 \log_x z$ are in $A.P.$,then

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