A) AP
B) GP
C) HP
D) None of these
Correct Answer: C
Solution :
Given progression are \[lo{{g}_{3}}2,lo{{g}_{6}}2,lo{{g}_{12}}2\] Now, \[\frac{1}{{{\log }_{6}}2}-\frac{1}{{{\log }_{3}}2}={{\log }_{2}}6-{{\log }_{2}}3\] \[={{\log }_{2}}\left( \frac{6}{3} \right)\] \[={{\log }_{2}}2=1\] \[\frac{1}{{{\log }_{12}}2}-\frac{1}{{{\log }_{6}}2}={{\log }_{2}}12-{{\log }_{2}}6\] \[={{\log }_{2}}\left( \frac{12}{6} \right)\] \[={{\log }_{2}}2=1\] \[\because \]\[\frac{1}{{{\log }_{6}}2}-\frac{1}{{{\log }_{3}}2}=\frac{1}{{{\log }_{12}}2}-\frac{1}{{{\log }_{6}}2}\] \[\therefore \] \[{{\log }_{3}}2,{{\log }_{6}}2,{{\log }_{12}}2\]are in HP.
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