If $\log _e \mathrm{a}, \log _e \mathrm{~b}, \log _e \mathrm{c}$ are in an A.P. and $$\log _e \mathrm{a}-\log _e 2 \mathrm{~b}, \log _e 2 \mathrm{~b}-\log _e 3 \mathrm{c}, \log _e 3 \mathrm{c} -\log _e$$ a are also in an A.P, then $a: b: c$ is equal to
Solution
<p>$$\log _{\mathrm{e}} \mathrm{a}, \log _{\mathrm{e}} \mathrm{b}, \log _{\mathrm{e}} \mathrm{c}$$ are in A.P.</p>
<p>$\therefore \mathrm{b}^2=\mathrm{ac}$ ..... (i)</p>
<p>Also</p>
<p>$$\begin{aligned}
& \log _e\left(\frac{a}{2 b}\right), \log _e\left(\frac{2 b}{3 c}\right), \log _e\left(\frac{3 c}{a}\right) \text { are in A.P. } \\
& \left(\frac{2 b}{3 c}\right)^2=\frac{a}{2 b} \times \frac{3 c}{a} \\
& \frac{b}{c}=\frac{3}{2}
\end{aligned}$$</p>
<p>Putting in eq. (i) $b^2=a \times \frac{2 b}{3}$</p>
<p>$$\begin{aligned}
& \frac{\mathrm{a}}{\mathrm{b}}=\frac{3}{2} \\
& \mathrm{a}: \mathrm{b}: \mathrm{c}=9: 6: 4
\end{aligned}$$</p>
About this question
Subject: Mathematics · Chapter: Sequences and Series · Topic: Arithmetic Progression
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