Answer
Verified
35.4k+ views
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.
Recently Updated Pages
To get a maximum current in an external resistance class 1 physics JEE_Main
f a body travels with constant acceleration which of class 1 physics JEE_Main
If the beams of electrons and protons move parallel class 1 physics JEE_Main
Let f be a twice differentiable such that fleft x rightfleft class 11 maths JEE_Main
Find the points of intersection of the tangents at class 11 maths JEE_Main
For the two circles x2+y216 and x2+y22y0 there isare class 11 maths JEE_Main
Other Pages
Explain the construction and working of a GeigerMuller class 12 physics JEE_Main
Differentiate between homogeneous and heterogeneous class 12 chemistry JEE_Main
If a wire of resistance R is stretched to double of class 12 physics JEE_Main
A gas is compressed isothermally to half its initial class 11 physics JEE_Main
A closed organ pipe and an open organ pipe are tuned class 11 physics JEE_Main
Oxidation state of S in H2S2O8 is A 6 B 7 C +8 D 0 class 12 chemistry JEE_Main