Question

The value of $\sum _ { r = 1 } ^ { n } \log \left( \frac { a ^ { r } } { b ^ { r - 1 } } \right)$ 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. $\frac { n } { 2 } \log \left( \frac { a ^ { n + 1 } } { b ^ { n + 1 } } \right)$

Answer 1

Answer: (C)

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

Answer 2

The given series is

This is an A.P. with first term $\log a$ and the common difference $\log \left( \frac { a ^ { 2 } } { b } \right) - \log a = \log \left( \frac { a } { b } \right)$

Therefore the sum of $n$ 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 $n = 1,2$.