Mathematics · Sequence and Series

JEE Main 2024 — 29 January, Shift 2 — Question 9

If log⁡ea,log⁡eb,log⁡ec\log _{e} a, \log _{e} b, \log _{e} c are in an A.P. and log⁡ea−\log _{e} a- log⁡e2b,log⁡e2b−log⁡e3c,log⁡e3c−log⁡e\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\mathrm{a}: \mathrm{b}: \mathrm{c} is equal to

  1. Option A:

    9:6:49: 6: 4

    Correct
  2. Option B:

    16:4:116: 4: 1

  3. Option C:

    25:10:425: 10: 4

  4. Option D:

    6:3:26: 3: 2

Answer: A

Step-by-step solution

log⁡ea,log⁡eb,log⁡ec\log _{\mathrm{e}} \mathrm{a}, \log _{\mathrm{e}} \mathrm{b}, \log _{\mathrm{e}} \mathrm{c} are in A.P.

∴b2=ac\therefore \mathrm{b}^{2}=\mathrm{ac} Also log⁡e(a2b),log⁡e(2b3c),log⁡e(3ca)\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.

(2b3c)2=a2 b×3ca\left(\frac{2 b}{3 c}\right)^{2}=\frac{\mathrm{a}}{2 \mathrm{~b}} \times \frac{3 \mathrm{c}}{\mathrm{a}}

bc=32\frac{\mathrm{b}}{\mathrm{c}}=\frac{3}{2}

Putting in eq. (i) b2=a×2 b3\mathrm{b}^{2}=\mathrm{a} \times \frac{2 \mathrm{~b}}{3}

ab=32\begin{aligned}& \frac{a}{b}=\frac{3}{2} &\end{aligned}

a:b:c=9:6:4 a: b: c=9: 6: 4

Answer key and solution verified before publishing.

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Exam
JEE Main 2024
Subject
Mathematics
Chapter
Sequence and Series
Topic
Arithmetic Progression
If log e a, log e b, log e c are in an A.P. and log e a- log e 2 b… | JEE Main 2024 PYQ with Solution · DhiX AI