MCQ
If $\log _e \mathrm{a}, \log _e \mathrm{~b}, \log _e \mathrm{c}$ are in an $A.P.$ and $\log _e \mathrm{a}-$ $\log _e 2 b, \log _e 2 b-\log _e 3 c, \log _e 3 c-\log _e a$ are also in an $A.P,$ then $a: b: c$ is equal to
  •  $9: 6: 4$
  • B
     $16: 4: 1$
  • C
     $25: 10: 4$
  • D
     $6: 3: 2$

Answer

Correct option: A.
 $9: 6: 4$
a
$\log _e a, \log _e b, \log _e c$ are in $ A.P.$

$\therefore \mathrm{b}^2=\mathrm{ac}$

Also

$\log _{\circ}\left(\frac{a}{2 b}\right), \log _{\circ}\left(\frac{2 b}{3 c}\right), \log _{\circ}\left(\frac{3 c}{a}\right)$ are in $A.P.$

$\left(\frac{2 b}{3 \mathrm{c}}\right)^2=\frac{\mathrm{a}}{2 \mathrm{~b}} \times \frac{3 \mathrm{c}}{\mathrm{a}} $

$ \frac{\mathrm{b}}{\mathrm{c}}=\frac{3}{2}$

Putting in eq. $(i)$ $b^2=a \times \frac{2 b}{3}$

$ \frac{a}{b}=\frac{3}{2}$

$ a: b: c=9: 6: 4$

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