A) \[{{\tan }^{-1}}\left( \frac{5}{1+{{a}_{n}}{{a}_{n-1}}} \right)\]
B) \[{{\tan }^{-1}}\left( \frac{5{{a}_{1}}}{1+{{a}_{n}}{{a}_{1}}} \right)\]
C) \[{{\tan }^{-1}}\left( \frac{5n-5}{1+{{a}_{n}}{{a}_{1}}} \right)\]
D) \[{{\tan }^{-1}}\left( \frac{5n-5}{1+{{a}_{n}}{{a}_{n+1}}} \right)\]
E) \[{{\tan }^{-1}}\left( \frac{5n}{1+{{a}_{1}}{{a}_{n}}} \right)\]
Correct Answer: C
Solution :
We have \[{{\tan }^{-1}}\left( \frac{5}{1+{{a}_{1}}{{a}_{2}}} \right)+{{\tan }^{-1}}\left( \frac{5}{1+{{a}_{2}}{{a}_{3}}} \right)+....\] \[+{{\tan }^{-1}}\left( \frac{5}{1+{{a}_{n-1}}{{a}_{n}}} \right)\] \[={{\tan }^{-1}}\left( \frac{{{a}_{2}}-{{a}_{1}}}{1+{{a}_{1}}{{a}_{2}}} \right)+{{\tan }^{-1}}\left( \frac{{{a}_{3}}-{{a}_{2}}}{1+{{a}_{2}}{{a}_{3}}} \right)+\] \[....+{{\tan }^{-1}}\left( \frac{{{a}_{n}}-{{a}_{n-1}}}{1+{{a}_{n-1}}{{a}_{n}}} \right)\] \[\left( \begin{align} & \because {{a}_{1}},{{a}_{2}},{{a}_{3}},......{{a}_{n}}\,are\text{ }in\text{ }AP\text{ }with\text{ }common \\ & difference\text{ }5\text{ }and\,{{a}_{i}}{{a}_{j}}\ne -1,i,j=1,2,....n \\ \end{align} \right)\] \[=({{\tan }^{-1}}{{a}_{2}}-{{\tan }^{-1}}{{a}_{1}})+({{\tan }^{-1}}{{a}_{3}}-{{\tan }^{-1}}{{a}_{2}})\] \[+....+({{\tan }^{-1}}{{a}_{n}}-{{\tan }^{-1}}{{a}_{n-1}})\] \[={{\tan }^{-1}}{{a}_{n}}-{{\tan }^{-1}}{{a}_{1}}\] \[={{\tan }^{-1}}\left( \frac{{{a}_{n}}-{{a}_{1}}}{1+{{a}_{n}}{{a}_{1}}} \right)\] \[={{\tan }^{-1}}\left( \frac{(n-1)5}{1+{{a}_{n}}{{a}_{1}}} \right)\] \[={{\tan }^{-1}}\left( \frac{5n-5}{1+{{a}_{n}}{{a}_{1}}} \right)\]
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