Question

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

• A. $z^3=y^{-1}=x$
• B. $x=y^{-2}$
• C. $2y=z^{-3}$
• D. $x=y^{-2}=z^4$

Answer 1

Answer: (A)

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} $
and $-15 log_{x} z = 1+3d $
$\Rightarrow z=x ^{ -\frac{\left(1+3d\right)}{15}}$
$ \therefore x= y^{1+d} = z^{\left(1+2d\right)\left(1+d\right)} $
$ = x^{-\left(1+2d\right)\left(1+d\right) \frac{\left(1+3d\right)}{15}} $
$ \therefore \left(1+d\right)\left(1+2d\right)\left(1+3d\right) = -15 $
$ \Rightarrow 6d^{3} +11d^{2}+6d +16 = 0 $
$ \Rightarrow \left(d+2\right)\left(6d^{2}-d+8\right)= 0$
$ \Rightarrow d= -2 $
[ $\because 6d^{2} -d+8=0$ does not give a real root]
$ \therefore x = y^{\frac{1}{2}} = y^{-1} $
$ = y =z^{1-4} = z^{-3} $
$z= x^{\frac{1}{3}}$
$ \therefore x= z^{3} $
Thus $x= y^{-1}=z^{3}$