$\log_{3} 2$ , $\log_{6} 2$ , $\log_{12} 2$ are in

A.P

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G.P

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H.P

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None of these

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Solution

The correct option is C
H.P

Solve the given series:
We know that $\log_{a} b = \frac{1}{\log_{b} a}$
$x = \log_{3} 2 \Rightarrow \frac{1}{x} = \log_{2} 3$
$y = \log_{6} 2 \Rightarrow \frac{1}{y} = \log_{2} 6 = \log_{2} (2 \times 3) = \log_{2} 2 + \log_{2} 3 [∵ \log_{a} (b \times c) = \log_{a} b + \log_{a} c] = 1 + \log_{2} 3 [∵ \log_{a} a = 1]$
$z = \log_{12} 2 \Rightarrow \frac{1}{z} = \log_{2} 12 = \log_{2} (2 \times 2 \times 3) = \log_{2} 2 + \log_{2} 2 + \log_{2} 3 [∵ \log_{a} (b \times c) = \log_{a} b + \log_{a} c] = 1 + 1 + \log_{2} 3 [∵ \log_{a} a = 1] = 2 + \log_{2} 3$
Clearly $\frac{1}{x}, \frac{1}{y}, \frac{1}{z}$ are in arithmetic progression, since the first term is $\log_{3} 2$ and the common difference is $1$ .
Therefore $x, y, z$ are in H.P. Since H.P is the reciprocal of arithmetic progression.
Hence, the correct answer is option (C).

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