Question

If ${\log }_x\left(ax\right),{\log }_x\left(bx\right)$ and ${\log }_x\left(cx\right)$ are in AP, where $a,b,c$ and $x$ belong to $(1,\infty )$, then $a,b$ and $c$ are in

• A. A.P.
• B. H.P.
• C. G.P.
• D. None of these

Answer 1

The correct option is C

G.P.

Explanation for the correct option:

Given, ${\log }_x\left(ax\right),{\log }_x\left(bx\right)$ and ${\log }_x\left(cx\right)$ are in A.P.

So, $\left({\log }_x\left(a\right)+{\log }_x\left(x\right)\right),\left({\log }_x\left(b\right)+{\log }_x\left(x\right)\right)$ and $\left({\log }_x\left(c\right)+{\log }_x\left(x\right)\right)$ are in A.P. $\left[\because \log \left(\mathrm{ab}\right)=\log \left(a\right)+\log \left(b\right)\right]$

We know that if $p,q,r$ are in A.P. then $2q=p+r$

So,

$$\begin{gathered} \Rightarrow 2\left({\log }_x\left(b\right)+1\right)=({\log }_x\left(a\right)+1)+\left({\log }_x\left(c\right)+1\right)\mathbf{\left[}\mathbf{\because }\mathbf{l}\mathbf{o}{\mathbf{g}}_{\mathbf{x}}\mathbf{\left(}\mathbf{x}\mathbf{\right)}\mathbf{=}\mathbf{1}\mathbf{\right]} \\ \Rightarrow 2{\log }_x\left(b\right)+\cancel{2}={\log }_x\left(a\right)+{\log }_x\left(c\right)+\cancel{2} \\ \Rightarrow 2\frac{\log \left(b\right)}{\log \left(x\right)}=\frac{\log \left(a\right)}{\log \left(x\right)}+\frac{\log \left(c\right)}{\log \left(x\right)}\mathbf{}\mathbf{}\mathbf{[}\mathbf{\because }{\mathbf{log}}_{\mathbf{x}}\left(y\right)\mathbf{=}\frac{\mathbf{log}\left(y\right)}{\mathbf{log}\left(x\right)}] \\ \Rightarrow \log {\left(b\right)}^2=\log \left(ac\right)\mathbf{\left[}\mathbf{\because }\mathbf{a}\mathbf{.}\mathbf{log}\mathbf{\left(}\mathbf{x}\mathbf{\right)}\mathbf{=}\mathbf{log}{\left(x\right)}^{\mathbf{a}}\mathbf{,}\mathbf{}\mathbf{log}\left(x\right)\mathbf{+}\mathbf{log}\left(y\right)\mathbf{=}\mathbf{log}\mathbf{\left(}\mathbf{xy}\mathbf{\right)}\mathbf{\right]} \\ \Rightarrow b^2=ac \end{gathered}$$

Since, $a,b,c$ satisfy the condition for a geometric progression, thus, they are in a G.P.

Hence, option(C) i.e. G.P. is the correct answer.