If log e a , log e b , log e c are in an A.P. and logea - loge2b , loge2b - log e 3 c , log e 3 c - log e a are also in an A.P., then a : b : c is equal to [2024]

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Solution

Q.
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 [2024]

1 16 : 4 : 1

2 9 : 6 : 4

3 6 : 3 : 2

4 25 : 10 : 4

Ans.
(2)

We have, $\log_{e} a, \log_{e} b, \log_{e} c$ are in an A.P.

$\Rightarrow 2 \log_{e} b = \log_{e} a + \log_{e} c \Rightarrow b^{2} = a c$ ...(i)

Also, $\log_{e} a - \log_{e} 2 b, \log_{e} 2 b - \log_{e} 3 c, \log_{e} 3 c - \log_{e} a$ are in an A.P.

$\Rightarrow 2 (\log_{e} 2 b - \log_{e} 3 c) = \log_{e} a - \log_{e} 2 b + \log_{e} 3 c - \log_{e} a$

$\Rightarrow \log_{e} {(\frac{2 b}{3 c})}^{2} = \log_{e} (\frac{3 c}{2 b}) \Rightarrow \frac{4 b^{2}}{9 c^{2}} = \frac{3 c}{2 b}$

$\Rightarrow \frac{4 b^{2}}{9 c^{2}} = \frac{3 c}{2 b}$ ...(ii)

From (i) and (ii), we have

$a = \frac{9 c^{2}}{4 c} \Rightarrow a = \frac{9}{4} c$ $∴$ $a : b : c = \frac{9}{4} : \frac{3}{2} : 1 = 9 : 6 : 4$

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