JEE MAIN - Mathematics (2024 - 29th January Evening Shift - No. 11)

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
$$6: 3: 2$$
$$9: 6: 4$$
$$25: 10: 4$$
$$16: 4: 1$$

Explanation

$$\log _{\mathrm{e}} \mathrm{a}, \log _{\mathrm{e}} \mathrm{b}, \log _{\mathrm{e}} \mathrm{c}$$ are in A.P.

$$\therefore \mathrm{b}^2=\mathrm{ac}$$ ..... (i)

Also

$$\begin{aligned} & \log _e\left(\frac{a}{2 b}\right), \log _e\left(\frac{2 b}{3 c}\right), \log _e\left(\frac{3 c}{a}\right) \text { are in A.P. } \\ & \left(\frac{2 b}{3 c}\right)^2=\frac{a}{2 b} \times \frac{3 c}{a} \\ & \frac{b}{c}=\frac{3}{2} \end{aligned}$$

Putting in eq. (i) $$b^2=a \times \frac{2 b}{3}$$

$$\begin{aligned} & \frac{\mathrm{a}}{\mathrm{b}}=\frac{3}{2} \\ & \mathrm{a}: \mathrm{b}: \mathrm{c}=9: 6: 4 \end{aligned}$$

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