Question

${\log }_{3}2,{\log }_{6}2,{\log }_{12}2$ are in

• A. HP
• B. AP
• C. GP
• D. None of these

Answer 1

Answer: (A)

HP

We have, ${\log }_{3}2,{\log }_{6}2,{\log }_{12}2$

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