The terms log _3 2, log _6 2, log _{12} 2 are | vedclass

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(C) Let the given terms be $a = \log _3 2, b = \log _6 2, c = \log _{12} 2$.Taking the reciprocals of these terms:$\frac{1}{a} = \log _2 3$$\frac{1}{b

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$H.P.$

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(C) Let the given terms be $a = \log _3 2, b = \log _6 2, c = \log _{12} 2$.Taking the reciprocals of these terms:$\frac{1}{a} = \log _2 3$$\frac{1}{b} = \log _2 6 = \log _2 (2 	imes 3) = \log _2 2 + \log _2 3 = 1 + \log _2 3$$\frac{1}{c} = \log _2 12 = \log _2 (4 	imes 3) = \log _2 4 + \log _2 3 = 2 + \log _2 3$Let $x = \log _2 3$. Then the reciprocals are $x, 1+x, 2+x$.Since the difference between consecutive terms is constant $(1)$,these reciprocals are in $A.P.$Therefore,the original terms are in $H.P.$

The terms $\log _3 2, \log _6 2, \log _{12} 2$ are in:

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