A) \[\frac{(n-1)d}{{{a}_{1}}+{{a}_{n}}}\]
B) \[\frac{(n-1)d}{1+{{a}_{1}}{{a}_{n}}}\]
C) \[\frac{nd}{1+{{a}_{1}}{{a}_{n}}}\]
D) \[\frac{{{a}_{n}}-{{a}_{1}}}{{{a}_{n}}+{{a}_{1}}}\]
Correct Answer: B
Solution :
| [b] We have, |
| \[{{\tan }^{-1}}\left( \frac{d}{1+{{a}_{1}}{{a}_{2}}} \right)+{{\tan }^{-1}}\left( \frac{d}{1+{{a}_{2}}{{a}_{3}}} \right)+...\] |
| \[+{{\tan }^{-1}}\left( \frac{d}{1+{{a}_{n-1}}{{a}_{n}}} \right)\] |
| \[={{\tan }^{-1}}\left( \frac{{{a}_{2}}-{{a}_{1}}}{1+{{a}_{1}}{{a}_{2}}} \right)+{{\tan }^{-1}}\left( \frac{{{a}_{3}}-{{a}_{2}}}{1+{{a}_{2}}{{a}_{3}}} \right)+...\] |
| \[+{{\tan }^{-1}}\left( \frac{{{a}_{n}}-{{a}_{n-1}}}{1+{{a}_{n-1}}{{a}_{n}}} \right)\] |
| \[=\left( {{\tan }^{-1}}{{a}_{2}}-{{\tan }^{-1}}{{a}_{1}} \right)+\left( {{\tan }^{-1}}{{a}_{3}}-{{\tan }^{-1}}{{a}_{2}} \right)+...\] \[+\left( {{\tan }^{-1}}{{a}_{n}}-{{\tan }^{-1}}{{a}_{n-1}} \right)\] |
| \[={{\tan }^{-1}}{{a}_{n}}-{{\tan }^{-1}}{{a}_{1}}={{\tan }^{-1}}\left( \frac{{{a}_{n}}-{{a}_{1}}}{1+{{a}_{n}}{{a}_{1}}} \right)\] |
| \[={{\tan }^{-1}}\left( \frac{(n-1)d}{1+{{a}_{1}}{{a}_{n}}} \right)\] \[\therefore \tan \left[ {{\tan }^{-1}}\left( \frac{d}{1+{{a}_{1}}{{a}_{2}}} \right)+{{\tan }^{-1}}\left( \frac{d}{1+{{a}_{2}}{{a}_{3}}} \right)+... \right.\] |
| \[\left. ...+{{\tan }^{-1}}\left( \frac{d}{1+{{a}_{n-1}}{{a}_{n}}} \right) \right]=\frac{(n-1d)}{1+{{a}_{1}}{{a}_{n}}}\] |
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