The numbers log_{3}{2}, log_{6}{2} and log_{1 | vedclass

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(C) Let the given numbers be $a = log_{3}{2}$,$b = log_{6}{2}$,and $c = log_{12}{2}$.We know that $log_{x}{y} = \frac{1}{log_{y}{x}}$.Therefore,$\frac

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Harmonic Progression

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(C) Let the given numbers be $a = log_{3}{2}$,$b = log_{6}{2}$,and $c = log_{12}{2}$.We know that $log_{x}{y} = \frac{1}{log_{y}{x}}$.Therefore,$\frac{1}{a} = log_{2}{3}$,$\frac{1}{b} = log_{2}{6}$,and $\frac{1}{c} = log_{2}{12}$.Now,calculate the differences of the reciprocals:$\frac{1}{b} - \frac{1}{a} = log_{2}{6} - log_{2}{3} = log_{2}(\frac{6}{3}) = log_{2}{2} = 1$.$\frac{1}{c} - \frac{1}{b} = log_{2}{12} - log_{2}{6} = log_{2}(\frac{12}{6}) = log_{2}{2} = 1$.Since the difference between consecutive reciprocals is constant,the reciprocals are in Arithmetic Progression.Therefore,the original numbers are in Harmonic Progression.

The numbers $log_{3}{2}, log_{6}{2}$ and $log_{12}{2}$ are in which progression?

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