Question

If $a_1,a_2,a_3,...,a_n$ are in A.P. with common difference $d$ where $a_r>0,\forall r=1,2,3,...,n,$ then $\tan [{\tan }^{-1}\left(\frac{d}{1+a_1{a}_2}\right)+{\tan }^{-1}\left(\frac{d}{1+a_2{a}_3}\right)+\cdots +{\tan }^{-1}\left(\frac{d}{1+a_{n-1}{a}_n}\right)]=$

• 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 $\frac{(n-1)d}{1+a_1{a}_n}$

$\tan [{\tan }^{-1}\left(\frac{d}{1+a_1{a}_2}\right)+{\tan }^{-1}\left(\frac{d}{1+a_2{a}_3}\right)+\cdots +{\tan }^{-1}\left(\frac{d}{1+a_{n-1}{a}_n}\right)]$

$=\tan [{\tan }^{-1}\left(\frac{a_2-a_1}{1+a_1{a}_2}\right)+{\tan }^{-1}\left(\frac{a_3-a_2}{1+a_2{a}_3}\right)+\cdots +{\tan }^{-1}\left(\frac{a_n-a_{n-1}}{1+a_{n-1}{a}_n}\right)]$
$=\text{tan}({\text{tan}}^{-1}{a}_n-{\text{tan}}^{-1}{a}_1)$
$=\tan \left({\tan }^{-1}\left(\frac{a_n-a_1}{1+a_n{a}_1}\right)\right)$
$=\frac{a_n-a_1}{1+a_n{a}_1}=\frac{(n-1)d}{1+a_n{a}_1}[\because a_n=a_1+(n-1)d]$