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 $lo{g}_{y}x=1+d$
$\Rightarrow x=y^{1+d}$
$lo{g}_{z}y=1+2d$
$\Rightarrow y=z^{1+2d}$
and $-15lo{g}_{x}z=1+3d$
$\Rightarrow z=x^{-\frac{\left(1+3d\right)}{15}}$
$\therefore x=y^{1+d}=z^{\left(1+2d\right)\left(1+d\right)}$
$=x^{-\left(1+2d\right)\left(1+d\right)\frac{\left(1+3d\right)}{15}}$
$\therefore \left(1+d\right)\left(1+2d\right)\left(1+3d\right)=-15$
$\Rightarrow 6d^3+11d^2+6d+16=0$
$\Rightarrow \left(d+2\right)\left(6d^2-d+8\right)=0$
$\Rightarrow d=-2$
[ $\because 6d^2-d+8=0$ 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 $x=y^{-1}=z^3$