Question

# The given terms, ${\log _3}2$, ${\log _6}2$, ${\log _{12}}2$ are in${\text{A}}{\text{.}}$ HP${\text{B}}{\text{.}}$ AP${\text{C}}{\text{.}}$ GP${\text{D}}{\text{.}}$ None of these

Hint- Make the base of all logs same and perform specific operations to find the required answer. Doing by eliminating options in competitive exams will be better.

Let $a = {\log _3}2$, $b = {\log _6}2$ and $c = {\log _{12}}2$
Since, ${\log _n}m = \dfrac{{\log m}}{{\log n}}$
$\Rightarrow a = \dfrac{{\log 2}}{{\log 3}}$, $b = \dfrac{{\log 2}}{{\log 6}}$, $c = \dfrac{{\log 2}}{{\log 12}}$
Here, the reciprocal of the given numbers are given by $\dfrac{1}{a} = \dfrac{{\log 3}}{{\log 2}},\dfrac{1}{b} = \dfrac{{\log 6}}{{\log 2}},\dfrac{1}{c} = \dfrac{{\log 12}}{{\log 2}}$
Now let us find out $\dfrac{1}{a} + \dfrac{1}{c} = \dfrac{{\log 3}}{{\log 2}} + \dfrac{{\log 12}}{{\log 2}} = \dfrac{{\log 3 + \log 12}}{{\log 2}}$
As we know that $\log m + \log n = \log mn$
$\Rightarrow \dfrac{1}{a} + \dfrac{1}{c} = \dfrac{{\log \left( {3 \times 12} \right)}}{{\log 2}} = \dfrac{{\log 36}}{{\log 2}} = \dfrac{{\log \left( {{6^2}} \right)}}{{\log 2}}$
Also we know that ${\text{log}}\left( {{m^n}} \right) = n\log m$
$\Rightarrow \dfrac{1}{a} + \dfrac{1}{c} = \dfrac{{2\log 6}}{{\log 2}} = \dfrac{2}{b}$.
Since, the condition for three numbers i.e., $a,b,c$ to be in Harmonic progression is $\dfrac{1}{a} + \dfrac{1}{c} = \dfrac{2}{b}$.
Therefore, the given three numbers i.e., $a = {\log _3}2,{\text{ }}b = {\log _6}2,{\text{ }}c = {\log _{12}}2$ are clearly in HP.
Therefore, option A is correct.

Note- For three numbers to be in Harmonic Progression (HP), twice the reciprocal of the middle number should be equal to the sum of the reciprocal of the other two numbers.