Question

If $a_1,a_2,a_3....,a_n$ is an A.P. with common difference d then $tan\left[ta{n}^{-1}\right(\frac{d}{1+a_1{a}_2}\left)\right]+ta{m}^{-1}\left(\frac{d}{1+a_2{a}_3}\right)]+...+ta{n}^{-1}(\frac{d}{1+a_{n-1}{a}_n}\left)\right]=$

• 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. $\frac{a_n-a_1}{a_n+a_1}$

Answer 1

The correct option is B $\frac{(n-1)d}{1+a_1{a}_n}$

We have $ta{n}^{-1}\left(\frac{d}{1+a_1{a}_2}\right)]+ta{m}^{-1}(\frac{d}{1+a_2{a}_3}\left)\right]+...+ta{n}^{-1}\left(\frac{d}{1+a_{n-1}{a}_n}\right)]==ta{n}^{-1}(\frac{a_2-a_1}{1+a_1{a}_2})+ta{n}^{-1}(\frac{a_3-a_2}{1+a_2{a}_3})+\cdots +ta{n}^{-1}(\frac{a_n-a_{n-1}}{1+a_{n-1}{a}_n})=(ta{n}^{-1}{a}_2-ta{n}^{-1}{a}_1)+(ta{n}^{-1}{a}_3-ta{n}^{-1}{a}_2)+.....+(ta{n}^{-1}{a}_n-ta{n}^{-1}{a}_{n-1})=ta{n}^{-1}{a}_n-ta{n}^{-1}{a}_1=ta{n}^{-1}(\frac{a_n-a_1}{1+a_n{a}_1})=ta{n}^{-1}(\frac{(n-1)d}{1+a_1{a}_n}).$