A) AP
B) GP
C) HP
D) None of these
Correct Answer: C
Solution :
Let \[a={{\log }_{3}}2,b={{\log }_{6}}2,c={{\log }_{12}}2\] \[\Rightarrow \]\[a=\frac{1}{{{\log }_{2}}3},b=\frac{1}{{{\log }_{2}}6},c=\frac{1}{{{\log }_{2}}12}\] \[\therefore \] \[\frac{2}{b}=2.{{\log }_{2}}6={{\log }_{2}}36\] and\[\frac{1}{a}+\frac{1}{c}={{\log }_{2}}3+{{\log }_{2}}12={{\log }_{2}}36\] \[\therefore \] \[\frac{2}{b}=\frac{1}{a}+\frac{1}{c}\] \[\Rightarrow \]a, b, c are in HP.
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