Prove that the numbers $\dfrac{1}{\log_3 2}$, $\dfrac{1}{\log_6 2}$ and $\dfrac{1}{\log_{12} 2}$ form an arithmetic progression (A.P).
To prove that (TPT) $\;$ $\dfrac{1}{\log_3 2}$, $\;$ $\dfrac{1}{\log_6 2}$ $\;$ and $\;$ $\dfrac{1}{\log_{12} 2}$ $\;$ form an A.P.
i.e. $\;$ TPT $\;$ $\log_2 3$, $\;$ $\log_2 6$ $\;$ and $\;$ $\log_2 12$ $\;$ form an A.P.
i.e. $\;$ TPT $\;$ $\log_2 6 - \log_2 3 = \log_2 12 - \log_2 6$
[Three numbers $\; a, \; b, \; c \;$ are in A.P. if $\;$ $b - a = c - b$]
i.e. $\;$ TPT $\;$ $\log_2 \left(\dfrac{6}{3}\right) = \log_2 \left(\dfrac{12}{6}\right)$
i.e. $\;$ TPT $\;$ $\log_2 2 = \log_2 2$ $\;\;$ which is true.
i.e. $\;$ TPT $\;$ $1 = 1$ $\;$ which is true.
$\implies$ The three numbers are in A.P.