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

Answer: (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, $\frac{1}{a}+\frac{1}{c}=\frac{\log 3}{\log 2}+\frac{\log 12}{\log 2}$
$=\frac{\log 3+\log 12}{\log 2}=\frac{\log 36}{\log 2}$
$[\because \log (mn)=\log m+\log n]$
$=\frac{\log 6^2}{\log 2}=\frac{2\log 6}{\log 2}=\frac{2}{b}$
Hence, a, b and c are in HP