Question

Prove that the numbers$\frac{1}{{\log }_{3}2},\frac{1}{{\log }_{6}2}$ and $\frac{1}{{\log }_{12}2}$ form an arithmetic progression.

Answer 1

$\frac{1}{lo{g}_{6}2}-\frac{1}{lo{g}_{3}2}=\frac{1}{lo{g}_{12}2}-\frac{1}{lo{g}_{6}2}$

$lo{g}_{2}6-lo{g}_{2}3=lo{g}_{2}12-lo{g}_{2}6$

$lo{g}_{2}2=lo{g}_{2}2⟹\text{LHS=RHS}$

$\therefore$ Proved