If a_1, a_2, a_3, dots, a_n is an A.P. with c | vedclass

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(B) Given that $a_1, a_2, \dots, a_n$ is an $A.P.$ with common difference $d$,we have $a_{k+1} - a_k = d$ for all $k = 1, 2, \dots, n-1$.Using the ide

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$\frac{(n - 1)d}{1 + a_1 a_n}$

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(B) Given that $a_1, a_2, \dots, a_n$ is an $A.P.$ with common difference $d$,we have $a_{k+1} - a_k = d$ for all $k = 1, 2, \dots, n-1$.Using the identity $	an^{-1} x - 	an^{-1} y = 	an^{-1} \left( \frac{x - y}{1 + xy} ight)$,we can write each term as:$	an^{-1} \left( \frac{d}{1 + a_k a_{k+1}} ight) = 	an^{-1} \left( \frac{a_{k+1} - a_k}{1 + a_k a_{k+1}} ight) = 	an^{-1} a_{k+1} - 	an^{-1} a_k$.Summing these terms from $k = 1$ to $n-1$ gives a telescoping series:$\sum_{k=1}^{n-1} (	an^{-1} a_{k+1} - 	an^{-1} a_k) = (	an^{-1} a_2 - 	an^{-1} a_1) + (	an^{-1} a_3 - 	an^{-1} a_2) + \dots + (	an^{-1} a_n - 	an^{-1} a_{n-1})$.All intermediate terms cancel out,leaving $	an^{-1} a_n - 	an^{-1} a_1$.Thus,the expression becomes $	an \left( 	an^{-1} a_n - 	an^{-1} a_1 ight) = 	an \left( 	an^{-1} \left( \frac{a_n - a_1}{1 + a_n a_1} ight) ight) = \frac{a_n - a_1}{1 + a_1 a_n}$.Since $a_n = a_1 + (n-1)d$,we have $a_n - a_1 = (n-1)d$.Therefore,the result is $\frac{(n - 1)d}{1 + a_1 a_n}$.

If $a_1, a_2, a_3, \dots, a_n$ is an $A.P.$ with common difference $d$,then $\tan \left[ \tan^{-1} \left( \frac{d}{1 + a_1 a_2} \right) + \tan^{-1} \left( \frac{d}{1 + a_2 a_3} \right) + \dots + \tan^{-1} \left( \frac{d}{1 + a_{n-1} a_n} \right) \right] = $

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