Question

\[{\log _3}2,{\log _6}2,{\log _{12}}2\] are in
A) A.P
B) G.P
C) H.P
D) None

Answer 1

Hint: The logarithmic function is the inverse function of the exponential function given by the formula\[{\log _b}a = c \Leftrightarrow {b^c} = \log a\], where b is the base of the logarithmic function. The logarithm is the mathematical operation that tells how many times a number or base is multiplied by itself to reach another number. There are five basic properties of the logarithm, namely Product rule, Quotient rule, Change of base rule, power rule, and equality rule.
A series is a description of the operation of adding infinitely many quantities, one after the other, to a given starting quantity.
To solve this question, first, use the change of base rule of the logarithm to write the logarithm in \[\dfrac{a}{b}\] form and then study the nature of the series by finding their common difference or the common ratio.

Complete step by step answer:
Use the change the base property of logarithm given as \[{\log _a}x = \dfrac{{{{\log }_b}x}}{{{{\log }_b}a}}\] hence the terms can write as
\[
{\log _3}2,{\log _6}2,{\log _{12}}2 \\
\Rightarrow\dfrac{{\log 2}}{{\log 3}},\dfrac{{\log 2}}{{\log 6}},\dfrac{{\log 2}}{{\log 12}} \\
\Rightarrow\dfrac{{\log 2}}{{\log 3}},\dfrac{{\log 2}}{{\log \left( {2 \times 3} \right)}},\dfrac{{\log 2}}{{\log \left( {{2^2} \times 3} \right)}} - - - - (i) \\
 \]
Since the Product rule of logarithm says \[{\log _a}\left( {xy} \right) = {\log _a}x + {\log _a}y\]; hence the denominators of the equation (i) can be written as:
\[\dfrac{{\log 2}}{{\log 3}},\dfrac{{\log 2}}{{\log 2 + \log 3}},\dfrac{{\log 2}}{{\log {2^2} + \log 3}} - - - - (ii)\]
Now use the power rule of the logarithm \[{\log _a}{x^p} = p{\log _a}x\] to resolve \[\log {2^2} = 2\log 2\], equation (ii) can be re-written as:
\[\dfrac{{\log 2}}{{\log 3}},\dfrac{{\log 2}}{{\log 2 + \log 3}},\dfrac{{\log 2}}{{2\log 2 + \log 3}} - - - - (iii)\]
Now divide both the numerators and the denominators of the equation (iii) with \[\log 3\]; hence equation (iii) is written as:
\[
\Rightarrow\dfrac{{\log 2}}{{\log 3}},\dfrac{{\log 2}}{{\log 2 + \log 3}},\dfrac{{\log 2}}{{2\log 2 + \log 3}} \\
\Rightarrow\dfrac{{\left( {\dfrac{{\log 2}}{{\log 3}}} \right)}}{{\left( {\dfrac{{\log 3}}{{\log 3}}} \right)}},\dfrac{{\left( {\dfrac{{\log 2}}{{\log 3}}} \right)}}{{\left( {\dfrac{{\log 2 + \log 3}}{{\log 3}}} \right)}},\dfrac{{\left( {\dfrac{{\log 2}}{{\log 3}}} \right)}}{{\left( {\dfrac{{2\log 2 + \log 3}}{{\log 3}}} \right)}} \\
\Rightarrow\dfrac{{\log 2}}{{\log 3}},\dfrac{{\left( {\dfrac{{\log 2}}{{\log 3}}} \right)}}{{\left( {\dfrac{{\log 2}}{{\log 3}} + 1} \right)}},\dfrac{{\left( {\dfrac{{\log 2}}{{\log 3}}} \right)}}{{\left( {2\dfrac{{\log 2}}{{\log 3}} + 1} \right)}} - - - - (iv) \\
 \]
Let \[\dfrac{{\log 2}}{{\log 3}} = a\], equation (iv) can be written as:
\[a,\dfrac{a}{{a + 1}},\dfrac{a}{{2a + 1}} - - - - (v)\]
So, the sequence of the given series \[{\log _3}2,{\log _6}2,{\log _{12}}2\] is \[a,\dfrac{a}{{a + 1}},\dfrac{a}{{2a + 1}}\]
Now check by the options for the nature of the series
i) To check for the A.P series, find the common difference of the sequence
\[
\Rightarrow{d_1} = \dfrac{a}{{a + 1}} - a;{d_2} = \dfrac{a}{{2a + 1}} - \dfrac{a}{{a + 1}} \\
\Rightarrow{d_1} = \dfrac{{a - {a^2} - a}}{{a + 1}};{d_2} = \dfrac{{a\left( {a + 1} \right) - a\left( {2a + 1} \right)}}{{\left( {2a + 1} \right) \times \left( {a + 1} \right)}} = \dfrac{{{a^2} + a - 2{a^2} - a}}{{2{a^2} + 3a + 1}} \\
\Rightarrow{d_1} = \dfrac{{ - {a^2}}}{{a + 1}};{d_2} = \dfrac{{ - {a^2}}}{{2{a^2} + 3a + 1}} \\
 \]
The common differences are not equal ${d_1} \ne {d_2}$ hence it can be concluded that the series is not in A.P
ii) To check for the G.P series, find the common ratio of the sequence
\[
\Rightarrow{r_1} = \dfrac{{\left( {\dfrac{a}{{a + 1}}} \right)}}{a};{r_2} = \dfrac{{\left( {\dfrac{a}{{2a + 1}}} \right)}}{{\left( {\dfrac{a}{{a + 1}}} \right)}} \\
\Rightarrow = \dfrac{1}{{a + 1}};{r_2} = \dfrac{{2a + 1}}{{a + 1}} \\
 \]
The common ratios are not equal ${r_1} \ne {r_2}$ hence the series is not in G.P
iii) To check for the H.P series, find the reciprocal of terms \[a,\dfrac{a}{{a + 1}},\dfrac{a}{{2a + 1}}\], which is
\[\dfrac{1}{a},\dfrac{{a + 1}}{a},\dfrac{{2a + 1}}{a}\]
Now find the common difference of the sequence
\[
\Rightarrow{d_1} = \left( {\dfrac{{a + 1}}{a}} \right) - \left( {\dfrac{1}{a}} \right);{d_2} = \left( {\dfrac{{2a + 1}}{a}} \right) - \left( {\dfrac{{a + 1}}{a}} \right) \\
\Rightarrow{d_1} = \dfrac{{a + 1 - 1}}{a};{d_2} = \dfrac{{2a + 1 - a - 1}}{a} \\
\Rightarrow{d_1} = \dfrac{a}{a};{d_2} = \dfrac{a}{a} \\
\Rightarrow{d_1} = 1;{d_2} = 1 \\
 \]
Since common differences are equal ${d_1} = {d_2}$ hence, it can be concluded that the series is in H.P

Thus option (C) is correct.

Note: To determine the nature of the series, always try to find the common difference or common ratios of the sequence series. Students often make mistakes in computing the common difference and the ratios of the consequent terms, which results in the wrong answer.