Question

The sequence $\log a,\log \left( {\dfrac{{{a^2}}}{b}} \right),\log \left( {\dfrac{{{a^3}}}{{{b^2}}}} \right),...$ is:
$\left( 1 \right)$ A G.P.
$\left( 2 \right)$ An A.P.
$\left( 3 \right)$ A H.P.
$\left( 4 \right)$ Both a G.P. and a H.P.

Answer 1

In order to solve this question, first simplify each term of the given series $\log a,\log \left( {\dfrac{{{a^2}}}{b}} \right),\log \left( {\dfrac{{{a^3}}}{{{b^2}}}} \right),...$. Then, use the condition of A.P. series that the difference of two consecutive numbers of series is the same to check if the given series is in A.P. or not.

Complete step-by-step solution:
Since, the given series is $\log a,\log \left( {\dfrac{{{a^2}}}{b}} \right),\log \left( {\dfrac{{{a^3}}}{{{b^2}}}} \right),...$
Now, we will simplify each term of the series as,
First term,
$\Rightarrow {a_{\kern 1pt} } = \log a$
Second term,
$\Rightarrow {a_2} = \log \left( {\dfrac{{{a^2}}}{b}} \right)$
Here, we will use the rule of algorithm as $\log \dfrac{m}{n} = \log m - \log n$ to simplify the above step.
$\Rightarrow {a_2} = \log {a^2} - \log b$
Now, we will use the rule of algorithm as $\log {m^n} = n\log m$ to simplify the obtained expression as:
$\Rightarrow {a_2} = 2\log a - \log b$
Third term,
$\Rightarrow {a_3} = \log \left( {\dfrac{{{a^3}}}{{{b^2}}}} \right)$
Here, we will use the rule of algorithm as $\log \dfrac{m}{n} = \log m - \log n$ to simplify the above step.
$\Rightarrow {a_3} = \log {a^3} - \log {b^2}$
Now, we will use the rule of algorithm as $\log {m^n} = n\log m$ to simplify the obtained expression as:
$\Rightarrow {a_3} = 3\log a - 2\log b$
Similarly, we can simplify the further terms of the series as,
$\Rightarrow {a_4} = 4\log a - 3\log b$
$\Rightarrow {a_5} = 5\log a - 4\log b$
…
Since, we can clearly see that the next term is increased by $\log a - \log b$ from the previous term.
Here, we can check it by calculating the common differences of two consecutive terms.
$d = {a_2} - {a_1} = {a_3} - {a_2} = ...$
We will calculate first ${a_2} - {a_1}$ as,
$\Rightarrow {a_2} - {a_1} = 2\log a - \log b - \log a$
$\Rightarrow {a_2} - {a_1} = \log a - \log b$
Now, we will calculate ${a_3} - {a_2}$ as,
$\Rightarrow {a_3} - {a_2} = 3\log a - 2\log b - \left( {2\log a - \log b} \right)$
$\Rightarrow {a_3} - {a_2} = 3\log a - 2\log b - 2\log a + \log b$
$\Rightarrow {a_3} - {a_2} = \log a - \log b$
Hence, the given series is in A.P.

Note: An Arithmetic progression series is an order sequence by incrementing a fixed constant number in each term. In other words, the difference between two consecutive terms of an A.P. is the same as a constant.