Question

$\log_3\,2, \log_6\,2, \log_{12}\,2$ are in

• A. $A.P$
• B. $G.P$
• C. $H.P$
• D. None of the options

Answer 1

The Correct Option is C

We have, $\log _{3} 2, \log _{6} 2, \log _{12} 2$ Let $a =\log _{3} 2=\frac{\log 2}{\log 3} \left[\because \log _{m} n=\frac{\log n}{\log m}\right]$ $b =\log _{6} 2=\frac{\log 2}{\log 6}$ and $c=\log _{12} 2=\frac{\log 2}{\log 12}$ Now, $\frac{1}{a}+\frac{1}{c}=\frac{\log 3}{\log 2}+\frac{\log 12}{\log 2}$ $=\frac{\log 3+\log 12}{\log 2}=\frac{\log 36}{\log 2}$ $[\because \log (m n)=\log m+\log n]$ $=\frac{\log 6^{2}}{\log 2}=\frac{2 \log 6}{\log 2}=\frac{2}{b}$ Hence, a, b and c are in HP