If $a_1, a_2, a_3, \ldots, a_n$ is an arithmetic progression with common difference $d$, then evaluate the following expression.
$$\begin{aligned} \tan \left[\tan ^{-1}\left(\frac{d}{1+a_1 a_2}\right)+\tan ^{-1}\left(\frac{d}{1+a_2 a_3}\right)\right. & +\tan ^{-1}\left(\frac{d}{1+a_3 a_4}\right) \\ & \left.+\ldots+\tan ^{-1}\left(\frac{d}{1+a_{n-1} a_n}\right)\right] \end{aligned}$$
$$\begin{aligned} \text{We have,}\quad& a_1=a, a_2=a+d, a_3=a+2 d \\ \text{and}\quad & d=a_2-a_1=a_3-a_2=a_4-a_3=\ldots=a_n-a_{n-1} \end{aligned}$$
Given that, $$\tan \left[\tan ^{-1}\left(\frac{d}{1+a_1 a_2}\right)+\tan ^{-1}\left(\frac{d}{1+a_2 a_3}\right)\right. \left.+\tan ^{-1}\left(\frac{d}{1+a_3 a_4}\right)+\ldots+\tan ^{-1}\left(\frac{d}{1+a_{n-1} \cdot a_n}\right)\right]$$
$$\begin{aligned} & =\tan \left[\tan ^{-1} \frac{a_2-a_1}{1+a_2 \cdot a_1}+\tan ^{-1} \frac{a_3-a_2}{1+a_3 \cdot a_2}+\ldots+\tan ^{-1} \frac{a_n-a_{n-1}}{1+a_n \cdot a_{n-1}}\right] \\ & =\tan \left[\left(\tan ^{-1} a_2-\tan ^{-1} a_1\right)+\left(\tan ^{-1} a_3-\tan ^{-1} a_2\right)+\ldots+\left(\tan ^{-1} a_n-\tan ^{-1} a_{n-1}\right)\right] \\ & =\tan \left[\tan ^{-1} a_n-\tan ^{-1} a_1\right] \\ & =\tan \left[\tan ^{-1} \frac{a_n-a_1}{1+a_n \cdot a_1}\right] \quad\left[\because \tan ^{-1} x-\tan ^{-1} y=\tan ^{-1}\left(\frac{x-y}{1+x y}\right)\right] \\ & =\frac{a_n-a_1}{1+a_n \cdot a_1} \quad\left[\because \tan \left(\tan ^{-1} x\right)=x\right] \end{aligned}$$
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