question_answer

The value of   \[\sum\limits_{r=1}^{n}{\log \left( \frac{{{a}^{r}}}{{{b}^{r-1}}} \right)}\] is

A) \[\frac{n}{2}\log \left( \frac{{{a}^{n}}}{{{b}^{n}}} \right)\]

B) \[\frac{n}{2}\log \left( \frac{{{a}^{n+1}}}{{{b}^{n}}} \right)\]

C) \[\frac{n}{2}\log \left( \frac{{{a}^{n+1}}}{{{b}^{n-1}}} \right)\]

D) \[\frac{n}{2}\log \left( \frac{{{a}^{n+1}}}{{{b}^{n+1}}} \right)\]

Correct Answer: C

Solution :

The given series is \[\log a+\log \left( \frac{{{a}^{2}}}{b} \right)+\log \left( \frac{{{a}^{3}}}{{{b}^{2}}} \right)+\log \left( \frac{{{a}^{4}}}{{{b}^{3}}} \right)+......+\log \left( \frac{{{a}^{n}}}{{{b}^{n-1}}} \right)\] This is an A.P. with first term  and the common difference \[\log \left( \frac{{{a}^{2}}}{b} \right)-\log a=\log \left( \frac{a}{b} \right)\] Therefore the sum of  terms is \[\frac{n}{2}\left[ \log a+\log \left( \frac{{{a}^{n}}}{{{b}^{n-1}}} \right) \right]=\frac{n}{2}\log \left( \frac{{{a}^{n+1}}}{{{b}^{n-1}}} \right)\]. Trick: Check for.