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8. Sequences and Series
hard
If $\log _e \mathrm{a}, \log _e \mathrm{~b}, \log _e \mathrm{c}$ are in an $A.P.$ and $\log _e \mathrm{a}-$ $\log _e 2 b, \log _e 2 b-\log _e 3 c, \log _e 3 c-\log _e a$ are also in an $A.P,$ then $a: b: c$ is equal to
A
$9: 6: 4$
B
$16: 4: 1$
C
$25: 10: 4$
D
$6: 3: 2$
(JEE MAIN-2024)
Solution
$\log _e a, \log _e b, \log _e c$ are in $ A.P.$
$\therefore \mathrm{b}^2=\mathrm{ac}$
Also
$\log _{\circ}\left(\frac{a}{2 b}\right), \log _{\circ}\left(\frac{2 b}{3 c}\right), \log _{\circ}\left(\frac{3 c}{a}\right)$ are in $A.P.$
$\left(\frac{2 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$
Standard 11
Mathematics
