A) AP
B) GP
C) HP
D) None of these
Correct Answer: C
Solution :
Consider,\[{{\log }_{2}}3,lo{{g}_{2}}6\]and \[{{\log }_{2}}12.\] We have, \[{{\log }_{2}}6={{\log }_{2}}(2\times 3)=lo{{g}_{2}}2+{{\log }_{2}}3\] \[=1+{{\log }_{2}}3\] and \[{{\log }_{2}}12={{\log }_{2}}({{2}^{2}}\times 3)\] \[=2{{\log }_{2}}2+{{\log }_{2}}3\] \[=2+{{\log }_{2}}3\] \[\therefore \] \[{{\log }_{2}}3,\,1+{{\log }_{2}}\,3\]and \[2+{{\log }_{2}}3\]are in AP \[\Rightarrow \]\[{{\log }_{2}}3,lo{{g}_{2}}\,6,{{\log }_{2}}\,12\]are in AP. \[\Rightarrow \]\[{{\log }_{3}}2,lo{{g}_{6}}\,2,{{\log }_{12}}\,2\]are in HP.
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