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

Answer 1

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.