Question

$lo{g}_{3}2,lo{g}_{6}2,lo{g}_{12}2$ are in

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

Answer 1

Answer: (C)

Given progression are
$lo{g}_{3}2,lo{g}_{6}2,lo{g}_{12}2$
Now, $\frac{1}{{\log }_{6}2}-\frac{1}{{\log }_{3}2}={\log }_{2}6-{\log }_{2}3$ $={\log }_2\left(\frac{6}{3}\right)$
$={\log }_{2}2=1$
$\frac{1}{{\log }_{12}2}-\frac{1}{{\log }_{6}2}={\log }_{2}12-{\log }_{2}6$
$={\log }_2\left(\frac{12}{6}\right)$
$={\log }_{2}2=1$
$\because$ $\frac{1}{{\log }_{6}2}-\frac{1}{{\log }_{3}2}=\frac{1}{{\log }_{12}2}-\frac{1}{{\log }_{6}2}$
$\therefore$ ${\log }_{3}2,{\log }_{6}2,{\log }_{12}2$ are in HP.