If a1, a2, a3, …., an is an arithmetic progression

As a1, a2, a3, …., an is an arithmetic progression with common difference d.

d = a2 – a1 = a3 – a2 = a4 – a3 = …… = an – an-1

So,

$\tan ^{-1} \frac{d}{1+a_{1} a_{2}}=\tan ^{-1} \frac{a_{2}-a_{1}}{1+a_{1} a_{2}}=\tan ^{-1} a_{2}-\tan ^{-1} a_{1}$

Similarly $\tan ^{-1} \frac{d}{1+a_{2} a_{3}}=\tan ^{-1} \frac{a_{3}-a_{2}}{1+a_{2} a_{3}}=\tan ^{-1} a_{3}-\tan ^{-1} a_{2}$

$\tan ^{-1} \frac{d}{1+a_{n-1} a_{n}}=\tan ^{-1} \frac{a_{n}-a_{n-1}}{1+a_{n-1} a_{n}}=\tan ^{-1} a_{n}-\tan ^{-1} a_{n-1}$

$\therefore \quad \tan \left[\tan ^{-1}\left(\frac{d}{1+a_{\mathrm{L}} a_{2}}\right)+\tan ^{-1}\left(\frac{d}{1+a_{2} a_{3}}\right)\right.$ $\left.+\tan ^{-1}\left(\frac{d}{1+a_{3} a_{4}}\right)+\cdots+\tan ^{-1}\left(\frac{d}{1+a_{n-1} a_{n}}\right)\right]$

$=\tan \left[\left(\tan ^{-1} a_{2}-\tan ^{-1} a_{1}\right)+\left(\tan ^{-1} a_{3}-\tan ^{-1} a_{2}\right)\right.$ $\left.+\ldots+\left(\tan ^{-1} a_{n}-\tan ^{-1} a_{n-1}\right)\right]$

$=\tan \left[\tan ^{-1} a_{n}-\tan ^{-1} a_{1}\right]$

$=\tan \left[\tan ^{-1} \frac{a_{n}-a_{1}}{1+a_{n} a_{1}}\right]\left[\because \tan ^{-1} x-\tan ^{-1} y=\tan ^{-1}\left(\frac{x-y}{1+x y}\right)\right]$

$=\frac{a_{n}-a_{1}}{1+a_{n} a_{1}} \quad\left[\because \tan \left(\tan ^{-1} x\right)=x\right]$