# 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

Last updated date: 16th Mar 2023

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Answer

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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.

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.

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