If log _e a, log _e b, log _e c are in an A.P | vedclass

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(A) Given $\log _e a, \log _e b, \log _e c$ are in $A.P.$,so $2 \log _e b = \log _e a + \log _e c$,which implies $b^2 = ac$ $(i)$.Also,$\log _e \left(

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$9 : 6 : 4$

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(A) Given $\log _e a, \log _e b, \log _e c$ are in $A.P.$,so $2 \log _e b = \log _e a + \log _e c$,which implies $b^2 = ac$ $(i)$.Also,$\log _e \left(\frac{a}{2b}ight), \log _e \left(\frac{2b}{3c}ight), \log _e \left(\frac{3c}{a}ight)$ are in $A.P.$Therefore,$2 \log _e \left(\frac{2b}{3c}ight) = \log _e \left(\frac{a}{2b}ight) + \log _e \left(\frac{3c}{a}ight)$.Using logarithmic properties,$\left(\frac{2b}{3c}ight)^2 = \frac{a}{2b} 	imes \frac{3c}{a} = \frac{3c}{2b}$.$\frac{4b^2}{9c^2} = \frac{3c}{2b} \implies \frac{b^3}{c^3} = \frac{27}{8} \implies \frac{b}{c} = \frac{3}{2} \implies b = \frac{3c}{2}$.Substitute $b = \frac{3c}{2}$ into $b^2 = ac$: $\left(\frac{3c}{2}ight)^2 = ac \implies \frac{9c^2}{4} = ac \implies a = \frac{9c}{4}$.Thus,$a : b : c = \frac{9c}{4} : \frac{3c}{2} : c = \frac{9}{4} : \frac{6}{4} : \frac{4}{4} = 9 : 6 : 4$.

If $\log _e a, \log _e b, \log _e c$ are in an $A.P.$ and $\log _e a - \log _e 2b, \log _e 2b - \log _e 3c, \log _e 3c - \log _e a$ are also in an $A.P.$,then $a : b : c$ is equal to

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