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Question: Answered & Verified by Expert
$\mathrm{a}, \mathrm{b}, \mathrm{c}$ are in G.P. with $1 < \mathrm{a} < \mathrm{b} < \mathrm{n}$, and $\mathrm{n}>1$ is an integer. $\log _{\text {an }}$ $\mathrm{n}, \log _{\mathrm{b}} \mathrm{n}, \log _{\mathrm{c}} \mathrm{n}$ form a sequence. This sequence is which
one of the following ?
MathematicsSequences and SeriesNDANDA 2007 (Phase 1)
Options:
  • A Harmonic progression
  • B Arithmetic progression
  • C Geometric progression
  • D None of these
Solution:
1972 Upvotes Verified Answer
The correct answer is: Harmonic progression
If a, b, c are in G.P. then, $\mathrm{b}^{2}=\mathrm{ac} \Rightarrow \mathrm{b}=(\mathrm{ac})^{1 / 2} \quad \ldots .(1)$
Taking $\log _{\mathrm{n}}$ on both the sides of eq. (1).
$\log _{\mathrm{n}} \mathrm{b}=\frac{1}{2}\left[\left(\log _{\mathrm{n}}(\mathrm{ac})\right]=\frac{\log _{\mathrm{n}} \mathrm{a}+\log _{\mathrm{n}} \mathrm{c}}{2}\right.$
or, $\frac{\log _{n} a+\log _{n} c}{2}=\log _{n} b$
So, $\log _{\mathrm{n}} \mathrm{a}, \log _{\mathrm{n}} \mathrm{b}$ and $\log _{\mathrm{n}} \mathrm{c}$ are in $\mathrm{AP}$.
Hence, $\frac{1}{\log _{n} a}, \frac{1}{\log _{n} b}, \frac{1}{\log _{n} c}$ are in H.P.
$\log _{\mathrm{a}} \mathrm{n}=\frac{1}{\log _{\mathrm{n}} \mathrm{a}}$
$\log _{\mathrm{b}} \mathrm{n}=\frac{1}{\log _{\mathrm{n}} \mathrm{b}}$
$\log _{\mathrm{c}} \mathrm{n}=\frac{1}{\log _{\mathrm{n}} \mathrm{c}}$
i.e. $\log _{\mathrm{a}} \mathrm{n}, \log _{\mathrm{b}} \mathrm{n}$, and $\log _{\mathrm{c}} \mathrm{n}$ are in $\mathrm{HP}$.

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