Question

The terms given by ${\log _3}2,{\log _6}2,{\log _{12}}2$ are in:
$\left( 1 \right)$ A.P.
$\left( 2 \right)$ G.P.
$\left( 3 \right)$ H.P.
\[\left( 4 \right)\] None of these

Answer 1

Hint: In order to solve this question, we will make the base of the algorithm the same. Then,we will substitute the value of each term in the condition of the given series in the option and will simplify it by using the appropriate mathematical operation. After that, we will check if it satisfies the condition or not.

Complete step-by-step solution:
Since, the given series is ${\log _3}2,{\log _6}2,{\log _{12}}2$.
Now, we will make the base the same. To make the base same, we will use the formula ${\log _a}b = \dfrac{1}{{{{\log }_b}a}}$ as:
\[ \Rightarrow {\log _3}2 = \dfrac{1}{{{{\log }_2}3}}\]
\[ \Rightarrow {\log _6}2 = \dfrac{1}{{{{\log }_2}6}}\]
\[ \Rightarrow {\log _{12}}2 = \dfrac{1}{{{{\log }_2}12}}\]
Here, the series will be \[\dfrac{1}{{{{\log }_2}3}},\dfrac{1}{{{{\log }_2}6}},\dfrac{1}{{{{\log }_2}12}}\].
Now, we will do this series by using all the conditions of given options one by one.
Option A
If a series $a,b,c$ in A.P.:
$ \Rightarrow b = \dfrac{{a + c}}{2}$
Now, we will substitute $\dfrac{1}{{{{\log }_2}3}}$ for $a$, $\dfrac{1}{{{{\log }_2}6}}$ for $b$ and $\dfrac{1}{{{{\log }_2}12}}$ for $c$ in the above condition of A.P.
$ \Rightarrow \dfrac{1}{{{{\log }_2}6}} = \dfrac{{\dfrac{1}{{{{\log }_2}3}} + \dfrac{1}{{{{\log }_2}12}}}}{2}$
Here, we will use the method of addition of fraction to simplify the above step as,
$ \Rightarrow \dfrac{1}{{{{\log }_2}6}} = \dfrac{{\dfrac{{{{\log }_2}12 + {{\log }_2}3}}{{{{\log }_2}3 \cdot {{\log }_2}12}}}}{2}$
Now, we will use the law of algorithmic to solve above expression as,
$\begin{align}
&\Rightarrow \dfrac{1}{{{{\log }_2}6}} = \dfrac{{{{\log }_2}12 \cdot 3}}{{2{{\log }_2}3 \cdot {{\log }_2}12}} \\
&\Rightarrow \dfrac{1}{{{{\log }_2}6}} = \dfrac{{{{\log }_2}36}}{{2{{\log }_2}3 \cdot {{\log }_2}12}} \\
&\Rightarrow \dfrac{1}{{{{\log }_2}6}} = \dfrac{{{{\log }_2}{6^2}}}{{2{{\log }_2}3 \cdot {{\log }_2}12}} \\
&\Rightarrow \dfrac{1}{{{{\log }_2}6}} = \dfrac{{2{{\log }_2}6}}{{2{{\log }_2}3 \cdot {{\log }_2}12}} \\
\end{align} $
Here, we will cancel out the term $2$ from the numerator and denominator of the right side of the obtained expression.
$ \Rightarrow \dfrac{1}{{{{\log }_2}6}} = \dfrac{{{{\log }_2}6}}{{{{\log }_2}3 \cdot {{\log }_2}12}}$
Since, $L.H.S. \ne R.H.S.$ So, the series is not in A.P.
Option B
If a series $a,b,c$ in G.P.:
$ \Rightarrow {b^2} = ac$
Now, we will substitute $\dfrac{1}{{{{\log }_2}3}}$ for $a$, $\dfrac{1}{{{{\log }_2}6}}$ for $b$ and $\dfrac{1}{{{{\log }_2}12}}$ for $c$ in the above condition of G.P.
$ \Rightarrow {\left( {\dfrac{1}{{{{\log }_2}6}}} \right)^2} = \dfrac{1}{{{{\log }_2}3}} \cdot \dfrac{1}{{{{\log }_2}12}}$
Now, we will use the law of algorithmic to solve above expression as,
$\begin{align}
&\Rightarrow \dfrac{1}{{{{\log }_2}{6^2}}} = \dfrac{1}{{{{\log }_2}3 \cdot {{\log }_2}12}} \\
&\Rightarrow \dfrac{1}{{2{{\log }_2}6}} = \dfrac{1}{{{{\log }_2}3 \cdot {{\log }_2}\left( {4 \cdot 3} \right)}} \\
&\Rightarrow \dfrac{1}{{2{{\log }_2}\left( {2 \cdot 3} \right)}} = \dfrac{1}{{{{\log }_2}3\left( {{{\log }_2}4 + {{\log }_2}3} \right)}} \\
&\Rightarrow \dfrac{1}{{2\left( {{{\log }_2}2 + {{\log }_2}3} \right)}} = \dfrac{1}{{{{\log }_2}3\left( {{{\log }_2}4 + {{\log }_2}3} \right)}} \\
\end{align} $
Since, $L.H.S. \ne R.H.S.$ So, the series is not in G.P.
Option C
If a series $a,b,c$ in H.P.:
$ \Rightarrow b = \dfrac{{2ac}}{{a + c}}$
Now, we will substitute $\dfrac{1}{{{{\log }_2}3}}$ for $a$, $\dfrac{1}{{{{\log }_2}6}}$ for $b$ and $\dfrac{1}{{{{\log }_2}12}}$ for $c$ in the above condition of H.P.
$ \Rightarrow \dfrac{1}{{{{\log }_2}6}} = \dfrac{{2 \cdot \dfrac{1}{{{{\log }_2}3}} \cdot \dfrac{1}{{{{\log }_2}12}}}}{{\dfrac{1}{{{{\log }_2}3}} + \dfrac{1}{{{{\log }_2}12}}}}$
Here, we will use the method of addition of fraction to simplify the above step as,
$ \Rightarrow \dfrac{1}{{{{\log }_2}6}} = \dfrac{{2 \cdot \dfrac{1}{{{{\log }_2}3 \cdot {{\log }_2}12}}}}{{\dfrac{{{{\log }_2}12 + {{\log }_2}3}}{{{{\log }_2}3 \cdot {{\log }_2}12}}}}$
After that, cancel out the equal terms.
$ \Rightarrow \dfrac{1}{{{{\log }_2}6}} = \dfrac{2}{{{{\log }_2}12 + {{\log }_2}3}}$
Now, we will use the law of algorithmic to solve above expression as,
$\begin{align}
&\Rightarrow \dfrac{1}{{{{\log }_2}6}} = \dfrac{2}{{{{\log }_2}\left( {12 \cdot 3} \right)}} \\
&\Rightarrow \dfrac{1}{{{{\log }_2}6}} = \dfrac{2}{{{{\log }_2}36}} \\
& \Rightarrow \dfrac{1}{{{{\log }_2}6}} = \dfrac{2}{{{{\log }_2}{6^2}}} \\
& \Rightarrow \dfrac{1}{{{{\log }_2}6}} = \dfrac{2}{{2{{\log }_2}6}} \\
\end{align} $
Here, we will cancel out the term $2$ from the numerator and denominator of the right side of the obtained expression.
$ \Rightarrow \dfrac{1}{{{{\log }_2}6}} = \dfrac{1}{{{{\log }_2}6}}$
Since, $L.H.S. = R.H.S.$ Thus, the series is in H.P.
Hence, the right option is option C.

Note: Some rules of algorithmic function used in the solution as,
$ \Rightarrow {\log _m}n = \dfrac{1}{{{{\log }_n}m}}$
$ \Rightarrow \log \left( {mn} \right) = \log m + \log n$
$ \Rightarrow \log {m^n} = n\log m$
For solving this type of problem we should have a good knowledge of all three types of progression.