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 H.P.

${\log }_{2}6=1+{\log }_{2}3{\log }_{2}12=2+{\log }_{2}3$
So,
${\log }_{2}3,{\log }_{2}6,{\log }_{2}12$ are in A.P.
Therefore,
${\log }_{3}2,{\log }_{6}2,{\log }_{12}2$ will be in H.P.