Question

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

• A. $0$
• B. $1$
• C. $2$
• D. None of these

Answer 1

The correct option is C $2$

Given $\tan [{\tan }^{-1}\left(\frac{d}{1+a_1{a}_2}\right)+{\tan }^{-1}\left(\frac{d}{1+a_2{a}_3}\right)+...+{\tan }^{-1}\left(\frac{d}{1+a_{n-1}{a}_n}\right)]=\frac{(n-p)d}{q+a_1{a}_n}$ ... $\left(i\right)$

Consider,

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

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

$=\tan [({\tan }^{-1}{a}_2-{\tan }^{-1}{a}_1)+({\tan }^{-1}{a}_3-{\tan }^{-1}{a}_2)+...+({\tan }^{-1}{a}_n-{\tan }^{-1}{a}_{n-1})]$ ..... [Using ${\tan }^{-1}a-{\tan }^{-1}b$]

$=\tan [{\tan }^{-1}{a}_n-{\tan }^{-1}{a}_1]$

$=\tan \left[{\tan }^{-1}\left(\frac{a_n-a_1}{1+a_n{a}_1}\right)\right]$

$=\tan \left[{\tan }^{-1}\left(\frac{(n-1)d}{1+a_1{a}_n}\right)\right]$ ..... $[\because a_n=a_1+(n-1\left)d\right]$

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

Comparing above equation with $\left(i\right)$ we get $p=1$ and $q=1$

$\therefore p+q=2$