Question

If a, b, c are in G.P., prove that $\frac{1}{lo{g}_{a}m},\frac{1}{lo{g}_{b}m},\frac{1}{lo{g}_{c}m}$ are in A.P.,

Answer 1

Here, a, b, c are in G.P., so

$b^2=ac$

Now taking $lo{g}_m$ on both the sides

$\Rightarrow lo{g}_m(b{)}^2=lo{g}_m(ac)$

$\Rightarrow 2lo{g}_m\left(b\right)=lo{g}_m\left(a\right)+lo{g}_m\left(c\right)$

$\Rightarrow \frac{2}{lo{g}_{b}m}=\frac{1}{lo{g}_{a}m}+\frac{1}{lo{g}_{c}m}$

$\Rightarrow \frac{1}{lo{g}_{b}m}-\frac{1}{lo{g}_{a}m}=\frac{1}{lo{g}_{c}m}-\frac{1}{lo{g}_{b}m}$

$\therefore \frac{1}{lo{g}_{a}m},\frac{1}{lo{g}_{b}m},\frac{1}{lo{g}_{c}m}$ are in A.P.