Question

If $1,{\log }_y\left(x\right),{\log }_z\left(y\right),-15{\log }_x\left(z\right)$ are in A. P. then the common difference of this A. P. is

• A. $1$
• B. $2$
• C. $-2$
• D. $3$

Answer 1

The correct option is C

$-2$

Step-1. Define common difference in Arithmetic progression.

Given, $1,{\log }_y\left(x\right),{\log }_z\left(y\right),-15{\log }_x\left(z\right)$ are in A.P.

Let $\begin{array}{rcl}{\mathrm{a}}_1& =& 1\end{array}$

$\begin{array}{rcl}{\mathrm{a}}_2& =& {\log }_{\mathrm{y}}\left(\mathrm{x}\right)\end{array}$

$\begin{array}{rcl}{\mathrm{a}}_3& =& {\log }_{\mathrm{z}}\left(\mathrm{y}\right)\end{array}$

$\begin{array}{rcl}{\mathrm{a}}_4& =& -15{\log }_{\mathrm{x}}\left(\mathrm{z}\right)\end{array}$

Now,

$\begin{array}{rcl}{\log }_{\mathrm{y}}\left(\mathrm{x}\right)-1& =& \mathrm{d}\end{array}$ $\begin{array}{rcl}\mathbf{[}\mathbf{\because }\mathbf{}{\mathbf{a}}_{\mathbf{2}}\mathbf{-}{\mathbf{a}}_{\mathbf{1}}& \mathbf{=}& {\mathbf{a}}_{\mathbf{3}}\mathbf{-}{\mathbf{a}}_{\mathbf{2}}\mathbf{=}{\mathbf{a}}_{\mathbf{4}}\mathbf{-}{\mathbf{a}}_{\mathbf{3}}\mathbf{=}\mathbf{d}\mathbf{]}\end{array}$

$\Rightarrow$ $\begin{array}{rcl}{\log }_{\mathrm{z}}\left(\mathrm{y}\right)-{\log }_{\mathrm{y}}\left(\mathrm{x}\right)& =& \mathrm{d}\end{array}$

$\Rightarrow$$\begin{array}{rcl}-15{\log }_{\mathrm{x}}\left(\mathrm{z}\right)-{\log }_{\mathrm{z}}\left(\mathrm{y}\right)& =& \mathrm{d}\end{array}$

Step-2. Convert the logarithm function into the exponential function:

${\log }_{\mathrm{y}}\left(\mathrm{x}\right)-1=\mathrm{d}$

$\Rightarrow$ ${\log }_{\mathrm{y}}\left(\mathrm{x}\right)=1+d$

$\Rightarrow$ $x=y^{1+d}\_\_\_\_\_\_\_\_\left(1\right)$ $\left[\mathbf{\because }{\mathbf{log}}_{\mathbf{a}}\mathbf{\left(}\mathbf{x}\mathbf{\right)}\mathbf{=}\mathbf{y}\mathbf{\Rightarrow }\mathbf{x}\mathbf{=}{\mathbf{a}}^{\mathbf{y}}\right]$

${\log }_{\mathrm{z}}\left(\mathrm{y}\right)-{\log }_{\mathrm{y}}\left(\mathrm{x}\right)=\mathrm{d}$

$\Rightarrow$ ${\log }_{\mathrm{z}}\left(\mathrm{y}\right)={\log }_{\mathrm{y}}\left(\mathrm{x}\right)+\mathrm{d}$

$\Rightarrow$ ${\log }_{\mathrm{z}}\left(\mathrm{y}\right)=1+\mathrm{d}+\mathrm{d}$

$\Rightarrow$ ${\log }_{\mathrm{z}}\left(\mathrm{y}\right)=1+2\mathrm{d}$

$\Rightarrow$ $\mathrm{y}={\mathrm{z}}^{1+2\mathrm{d}}\mathrm{\_\_\_\_\_\_}\left(2\right)$

$-15{\log }_{\mathrm{x}}\left(\mathrm{z}\right)-{\log }_{\mathrm{z}}\left(\mathrm{y}\right)=\mathrm{d}$

$\Rightarrow$ $-15{\log }_{\mathrm{x}}\left(\mathrm{z}\right)={\log }_{\mathrm{z}}\left(\mathrm{y}\right)+\mathrm{d}$

$\Rightarrow$ $-15{\log }_{\mathrm{x}}\left(\mathrm{z}\right)=1+2\mathrm{d}+\mathrm{d}$

$\Rightarrow$ ${\log }_{\mathrm{x}}\left(\mathrm{z}\right)=\frac{1+3\mathrm{d}}{-15}$

$\Rightarrow$ $\mathrm{z}={\mathrm{x}}^{\left(\frac{1+3\mathrm{d}}{-15}\right)}\mathrm{\_\_\_\_\_\_\_}\left(3\right)$

Step-3. Convert all the exponents into $x$ form

$$\begin{gathered} x=y^{1+d} \\ ={\left(z^{1+2d}\right)}^{1+d} \\ =z^{\left(1+2d\right)\left(1+d\right)} \\ ={\left[x^{\left(\frac{-\left(1+3d\right)}{15}\right)}\right]}^{\left(1+2d\right)\left(1+d\right)} \\ =x^{\frac{-1}{15}\left(1+3d\right)\left(1+2d\right)\left(1+d\right)} \end{gathered}$$

Step-4.Calculate the value of $\text{'}d\text{'}$

$\begin{array}{rcl}{x}^1& =& x^{\frac{-1}{15}\left(1+3d\right)\left(1+2d\right)\left(1+d\right)}\end{array}$

$\Rightarrow$$\begin{array}{rcl}\frac{-1}{15}\left(1+3d\right)\left(1+2d\right)\left(1+d\right)& =& 1\end{array}$

$\Rightarrow$ $\begin{array}{rcl}\left(1+3d\right)\left(1+2d\right)\left(1+d\right)& =& -15\end{array}$

$\Rightarrow$ $\begin{array}{rcl}\left(6d^2+5d+1\right)\left(1+d\right)& =& -15\end{array}$

$\Rightarrow$ $\begin{array}{rcl}6d^3+11d^2+6d+1& =& -15\end{array}$

$\Rightarrow$ $\begin{array}{rcl}6d^3+11d^2+6d+16& =& 0\end{array}$

$\Rightarrow$ $\begin{array}{rcl}\left(d+2\right)\left(6d^2-d+8\right)& =& 0\end{array}$

$\Rightarrow$ $\begin{array}{rcl}d+2& =& 0\end{array}$

$\begin{array}{rcl}\mathbf{\therefore }\mathbf{d}& \mathbf{=}& \mathbf{-}\mathbf{2}\end{array}$

Hence, Option(C) is the correct answer.