Question

\[a,b,c\] are in G.P with \[11\] is an integer, \[{{\log }_{a}}n,{{\log }_{b}}n,{{\log }_{c}}n\] form a sequence. This sequence is which one of the following?
A. Harmonic progression
B. Arithmetic progression
C. Geometric progression
D. None of these

Answer 1

Hint: In this question, we have to find the type of the given series. Here, we have given a known series that is in geometric progression. If some terms are said to be in geometric progression, then their geometric mean is the middle term of the series which is equal to the square root of the first and last terms of the series. By using this, we can able to find the type of the given series.

Formula Used: If \[a,b,c\] are in geometric progression, then their geometric mean is
$b=\sqrt{ac}$
Where $b$ is the middle term, $a$ is the first term and $c$ is the last term
Some of the important logarithmic formulae:
$\begin{align}
  & {{\log }_{a}}b=\dfrac{1}{{{\log }_{b}}a} \\
 & {{\log }_{a}}bc={{\log }_{a}}b+{{\log }_{a}}c \\
 & {{\log }_{a}}\left( \dfrac{b}{c} \right)={{\log }_{a}}b-{{\log }_{a}}c \\
\end{align}$

Complete step by step solution: Given that, \[a,b,c\] are in geometric progression.
Then, their geometric mean is
$b=\sqrt{ac}={{\left( ac \right)}^{{}^{1}/{}_{2}}}$
Applying logarithm with base $n$ to the above expression, we get
${{\log }_{n}}b={{\log }_{n}}{{\left( ac \right)}^{{}^{1}/{}_{2}}}$
On simplifying,
\[\begin{align}
  & \Rightarrow {{\log }_{n}}b=\dfrac{1}{2}{{\log }_{n}}\left( ac \right) \\
 & \Rightarrow {{\log }_{n}}b=\dfrac{1}{2}\left[ {{\log }_{n}}a+{{\log }_{n}}c \right] \\
\end{align}\]
So, we can write this as
\[{{\log }_{n}}b=\dfrac{1}{2}\left[ {{\log }_{n}}a+{{\log }_{n}}c \right]\text{ }...(1)\]
Thus, from (1), we can say that, the terms \[{{\log }_{n}}a,{{\log }_{n}}b,{{\log }_{n}}c\] are in arithmetic progression.
Then, their reciprocals are said to be in harmonic progression.
I.e., \[\dfrac{1}{{{\log }_{n}}a},\dfrac{1}{{{\log }_{n}}b},\dfrac{1}{{{\log }_{n}}c}\] are in Harmonic progression.
Then, according to the logarithmic rule, we can write
\[\begin{align}
  & \dfrac{1}{{{\log }_{n}}a}={{\log }_{a}}n \\
 & \dfrac{1}{{{\log }_{n}}b}={{\log }_{b}}n \\
 & \dfrac{1}{{{\log }_{n}}c}={{\log }_{c}}n \\
\end{align}\]
Therefore, \[{{\log }_{a}}n,{{\log }_{b}}n,{{\log }_{c}}n\] are in H.P (Harmonic progression).

Option ‘A’ is correct

Note: Here we need to remember that, if any series of terms is said to be proven for the type of progression, we need to find their means and equate them with their middle term. If they are the same, then they are said to be in respective progression.