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…

If logea,logeb,logec are in an A.P. and logea-loge2b,loge2b-loge3c,loge3c-logea are also in an A.P., then a:b:c is equal to
  1. 9: 6: 4
  2. 16: 4: 1
  3. 25: 10: 4
  4. 6: 3: 2

Solution

Given: $\log a, \log b, \log c$ are in $A.P.$ $\Rightarrow 2\log b = \log a + \log c$ $\Rightarrow b^2 = ac$ And $\log a - \log 2b, \log 2b - \log 3c, \log 3c - \log a$ are in $A.P.$ $\Rightarrow 2\log \frac{2b}{3c} = \log \frac{a}{2b} + \log \frac{3c}{a}$ $\Rightarrow \left(\frac{2b}{3c}\right)^2 = \frac{a}{2b} \times \frac{3c}{a}$ $\Rightarrow \left(\frac{2b}{3c}\right)^2 = \frac{3c}{2b}$ $\Rightarrow 2b = 3c$ $\Rightarrow 4b^2 = 9c^2$ and $6b = 9c$ Now solving the equation (1) & (2) we get, $4ac = 9c^2$ $\Rightarrow 4a = 9c$ Then from equation (2) & (3) we get, $4a = 6b = 9c = k$ $\Rightarrow a = $\frac{k}{4}$, b = \frac{k}{6}$ & $c = \frac{k}{9}$ $\Rightarrow a : b : c = $\frac{1}{4}$ : $\frac{1}{6}$ : $\frac{1}{9}$ = 9 : 6 : 4$

Asked in: JEE Main 2024 (29 Jan Shift 2)

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