Question

If $\theta =ta{n}^{-1}\frac{d}{1+a_1{a}_2}+ta{n}^{-1}\frac{d}{1+a_2{a}_3}+\cdots +ta{n}^{-1}\frac{d}{1+a_{n-1}{a}_n}$, where $a_1,a_2,a_3,\cdots a_n$ are in A.P. with common difference d, then $tan\theta =$

• A. $\frac{(n-1)d}{1+a_1{a}_n}$
• B. $\frac{(n-1)d}{a_1+a_n}$
• C. $\frac{a_n-a_1}{a_n+a_1}$
• D. $\frac{nd}{1+a_1{a}_n}$

Answer 1

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

$\theta =ta{n}^{-1}\frac{a_2-a_1}{1+a_1{a}_2}+ta{n}^{-1}\frac{a_3-a_2}{1+a_2{a}_3}+\cdots +ta{n}^{-1}\frac{a_n-a_{n-1}}{1+a_{n-1}{a}_n}$
$=(ta{n}^{-1}{a}_2-ta{n}^{-1}{a}_1)+(ta{n}^{-1}{a}_3-ta{n}^{-1}{a}_2)+\cdots +ta{n}^{-1}\frac{a_n-a_{n-1}}{1+a_{n-1}{a}_n}+\cdots +(ta{n}^{-1}{a}_n-ta{n}^{-1}{a}_{n-1})$
$=ta{n}^{-1}{a}_n-ta{n}^{-1}{a}_1$
$=ta{n}^{-1}\frac{a_n-a_1}{1+a_1{a}_n}=ta{n}^{-1}\frac{(n-1)d}{1+a_1{a}_n}$
$\therefore tan\theta =\frac{(n-1)d}{1+a_1{a}_n}$