+1 vote
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The sequence $\dfrac{1}{\log_3 2}, \: \dfrac{1}{\log_6 2}, \: \dfrac{1}{\log_{12} 2}, \: \dfrac{1}{\log_{24} 2} \dots$ is in

1. Arithmetic progression (AP)
2. Geometric progression ( GP)
3. Harmonic progression (HP)
4. None of these

recategorized | 83 views

As $\frac{1}{log_nk} = log_kn$ the sequence can be written as $log_23,log_26,log_212$ and so on. According to AP formula a,b,c are in AP iff b-a = c-b. Now in the given series $log_26 - log_23 = log_2{\frac{6}{3}} = 1$ and  $log_212 - log_26 = log_2{\frac{12}{6}} = 1$ and so on. So the given sequence is in AP. Option A is correct.
by (231 points)

$\log_{a}{b} \times \log_{b}{a}=\log_{a}{a}=1;~[\because \log_{a}{b} \times \log_{b}{c} = \log_{a}{c}\mathrm{~and~}\log_{k}{k}=1]$

$\therefore \log_{a}{b}=\frac{1}{\log_{b}{a}}$

Again we know that $\log_{k}{(ab)}=\log_{k}{a}+\log_{k}{b}$ and $\log_{k}{(a^r)}=r\log_{k}{a}$.

So we can derive $\log_{a}{a^r}=r\log_{a}{a}=r\times 1=r$

Now the progression is

\begin{align} &~~~~\frac{1}{\log_{3}{2}},\frac{1}{\log_{6}{2}},\frac{1}{\log_{12}{2}},\frac{1}{\log_{24}{2}},\cdots \\ &\equiv \log_{2}{3},~\log_{2}{6},~\log_{2}{12},~\log_{2}{24},\cdots \\ &\equiv \log_{2}{3},~\log_{2}{(2\times3)},~\log_{2}{(2^2\times3)},~\log_{2}{(2^3\times3)},\cdots \\ &\equiv \log_{2}{3},~\log_{2}{2}+\log_{2}{3},~\log_{2}{2^2}+\log_{2}{3},~\log_{2}{2^3}+\log_{2}{3},\cdots \\ &\equiv \log_{2}{3},~1+\log_{2}{3},~2+\log_{2}{3},~3+\log_{2}{3},\cdots \end{align}

Clearly $\log_{2}{3},~1+\log_{2}{3},~2+\log_{2}{3},~3+\log_{2}{3},\cdots$ is an Arithmetic progression because the difference between two consecutive terms is $1$.

So the correct answer is A.

by Active (3.5k points)