If a, b, c are GP, then show that

Solution:

To prove: $\frac{1}{\log _{a} m}, \frac{1}{\log _{b} m^{\prime}}, \frac{1}{\log _{c} m}$ are in AP.

Given: a, b, c are in GP

Formula used: (i) $\frac{1}{\log _{a} m}=\log _{m} a=\frac{\log a}{\log m}$

As, a, b, c are in GP

$\Rightarrow \frac{b}{a}=\frac{c}{b}$

Taking log both side $\log \frac{b}{a}=\log \frac{c}{b}$

$\Rightarrow \log b-\log a=\log c-\log b$

$\Rightarrow 2 \log b=\log a+\log c$

Dividing by log m

$\Rightarrow 2\left(\frac{\log \mathrm{b}}{\log \mathrm{m}}\right)=\frac{\log \mathrm{a}}{\log \mathrm{m}}+\frac{\log \mathrm{c}}{\log \mathrm{m}}$

$\Rightarrow 2 \log _{m} b=\log _{m} a+\log _{m} c \quad\left(\right.$ As, $\left.\log _{m} a=\frac{\log a}{\log m}\right)$

$2\left(\frac{1}{\log _{b} m}\right)=\frac{1}{\log _{a} m}+\frac{1}{\log _{c} m}\left(\right.$ As $\left.\frac{1}{\log _{a} m}=\log _{m} a\right)$

Whenever any number a,b,c are in AP then 2b = a+c, considering this and the above

equation we can say that $\frac{1}{\log _{a} m}, \frac{1}{\log _{b} m}, \frac{1}{\log _{c} m}$ are in AP

Hence proved