Sequences And Series Question 10
Question 10 - 2024 (29 Jan Shift 2)
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 $\mathrm{a}: \mathrm{b}: \mathrm{c}$ is equal to
(1) $9: 6: 4$
(2) $16: 4: 1$
(3) $25: 10: 4$
(4) $6: 3: 2$
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Answer (1)
Solution
$\log _{e} a, \log _{e} b, \log _{e} c$ are in A.P.
$\therefore \mathrm{b}^{2}=\mathrm{ac} \ldots(i)$
Also
$\log _{e}\left(\frac{a}{2 b}\right), \log _{e}\left(\frac{2 b}{3 c}\right), \log _{e}\left(\frac{3 c}{a}\right)$ are in A.P.
$\left(\frac{2 \mathrm{~b}}{3 \mathrm{c}}\right)^{2}=\frac{\mathrm{a}}{2 \mathrm{~b}} \times \frac{3 \mathrm{c}}{\mathrm{a}}$
$\frac{\mathrm{b}}{\mathrm{c}}=\frac{3}{2}$
Putting in eq. (i) $b^{2}=a \times \frac{2 b}{3}$
$\frac{a}{b}=\frac{3}{2}$
$a: b: c=9: 6: 4$
