Sequence And Series Question 485
Question: The value of $ \sum\limits_{r=1}^{n}{\log ( \frac{a^{r}}{{b^{r-1}}} )} $ is
Options:
A) $ \frac{n}{2}\log ( \frac{a^{n}}{b^{n}} ) $
B) $ \frac{n}{2}\log ( \frac{{a^{n+1}}}{b^{n}} ) $
C) $ \frac{n}{2}\log ( \frac{{a^{n+1}}}{{b^{n-1}}} ) $
D) $ \frac{n}{2}\log ( \frac{{a^{n+1}}}{{b^{n+1}}} ) $
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Answer:
Correct Answer: C
Solution:
The given series is $ \log a+\log ( \frac{a^{2}}{b} )+\log ( \frac{a^{3}}{b^{2}} )+\log ( \frac{a^{4}}{b^{3}} )+……+\log ( \frac{a^{n}}{{b^{n-1}}} ) $ This is an A.P. with first term and the common difference $ \log ( \frac{a^{2}}{b} )-\log a=\log ( \frac{a}{b} ) $ Therefore the sum of terms is $ \frac{n}{2}[ \log a+\log ( \frac{a^{n}}{{b^{n-1}}} ) ]=\frac{n}{2}\log ( \frac{{a^{n+1}}}{{b^{n-1}}} ) $ . Trick: Check for.
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