A) A.P.
B) H.P.
C) G.P.
D) None of these
Correct Answer: B
Solution :
[b] When \[{{\log }_{3}}2,lo{{g}_{6}}2,lo{{g}_{12}}2\] be in H.P. Then, \[{{\log }_{2}}3,lo{{g}_{2}}6,lo{{g}_{2}}6\]be in A.P. \[\Rightarrow 2.1o{{g}_{2}}6=lo{{g}_{2}}3+lo{{g}_{2}}12\] \[lo{{g}_{2}}{{6}^{2}}=lo{{g}_{2}}\left( 3\times 12 \right)=lo{{g}_{2}}36\] Hence, option [b] is correct.
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