If a, b, c are in G.P., prove that

Question:

If abc are in G.P., prove that

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

Solution:

abc are in G.P.

$\therefore b^{2}=a c$

Now taking $\log _{m}$ on both the sides:

$\Rightarrow \log _{m}(b)^{2}=\log _{m}(a c)$

$\Rightarrow 2 \log _{m}(b)=\log _{m} a+\log _{m}(c)$

$\Rightarrow \frac{2}{\log _{b}(m)}=\frac{1}{\log _{a}(m)}+\frac{1}{\log _{c}(m)}$

Thus, $\frac{1}{\log _{\mathrm{a}}(\mathrm{m})}, \frac{1}{\log _{\mathrm{b}}(\mathrm{m})}$ and $\frac{1}{\log _{\mathrm{c}}(\mathrm{m})}$ are in A.P.

 

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