If $\log _e a, \log _e b, \log _e c$ are in an A.P. and $\log _e a-\log _e 2 b, \log _e 2 b-\log _e 3 c, \log _e 3 c-\log _e$ a are also in an A.P, then $a: b: c$ is equal to
Select the correct option:
A
$9: 6: 4$
B
$16: 4: 1$
C
$25: 10: 4$
D
$6: 3: 2$
✓ Correct! Well done.
✗ Incorrect. Try again or view the solution.
Solution
$$
\begin{aligned}
& \log _e a, \log _e b, \log _e c \text { are in A.P. } \\
& \therefore \mathrm{b}^2=\mathrm{ac}
\end{aligned}
$$
Also
$$
\begin{aligned}
& \log _e\left(\frac{a}{2 b}\right), \log _e\left(\frac{2 b}{3 c}\right), \log _e\left(\frac{3 c}{a}\right) \text { are in A.P. } \\
& \left(\frac{2 b}{3 c}\right)^2=\frac{a}{2 b} \times \frac{3 c}{a} \\
& \frac{\mathrm{~b}}{\mathrm{c}}=\frac{3}{2}
\end{aligned}
$$
Putting in eq. (i) $b^2=a \times \frac{2 b}{3}$
$$
\begin{aligned}
& \frac{a}{b}=\frac{3}{2} \\
& a: b: c=9: 6: 4
\end{aligned}
$$
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