$\log_{a}{b} \times \log_{b}{a}=\log_{a}{a}=1;~[\because \log_{a}{b} \times \log_{b}{c} = \log_{a}{c}\mathrm{~and~}\log_{k}{k}=1]$
$\therefore \log_{a}{b}=\frac{1}{\log_{b}{a}}$
Again we know that $\log_{k}{(ab)}=\log_{k}{a}+\log_{k}{b}$ and $\log_{k}{(a^r)}=r\log_{k}{a}$.
So we can derive $\log_{a}{a^r}=r\log_{a}{a}=r\times 1=r$
Now the progression is
$\begin{align} &~~~~\frac{1}{\log_{3}{2}},\frac{1}{\log_{6}{2}},\frac{1}{\log_{12}{2}},\frac{1}{\log_{24}{2}},\cdots \\ &\equiv \log_{2}{3},~\log_{2}{6},~\log_{2}{12},~\log_{2}{24},\cdots \\ &\equiv \log_{2}{3},~\log_{2}{(2\times3)},~\log_{2}{(2^2\times3)},~\log_{2}{(2^3\times3)},\cdots \\ &\equiv \log_{2}{3},~\log_{2}{2}+\log_{2}{3},~\log_{2}{2^2}+\log_{2}{3},~\log_{2}{2^3}+\log_{2}{3},\cdots \\ &\equiv \log_{2}{3},~1+\log_{2}{3},~2+\log_{2}{3},~3+\log_{2}{3},\cdots \end{align}$
Clearly $\log_{2}{3},~1+\log_{2}{3},~2+\log_{2}{3},~3+\log_{2}{3},\cdots$ is an Arithmetic progression because the difference between two consecutive terms is $1$.
So the correct answer is A.