If a_1, a_2, dots, a_n are in A.P. with commo | vedclass

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(B) Given that $a_1, a_2, \dots, a_n$ are in $A.P.$ with common difference $d = a_2 - a_1 = a_3 - a_2 = \dots = a_n - a_{n-1}$.The given series is $S

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$\cot a_1 - \cot a_n$

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(B) Given that $a_1, a_2, \dots, a_n$ are in $A.P.$ with common difference $d = a_2 - a_1 = a_3 - a_2 = \dots = a_n - a_{n-1}$.The given series is $S = \sin d (\csc a_1 \csc a_2 + \csc a_2 \csc a_3 + \dots + \csc a_{n-1} \csc a_n)$.We can write $\sin d = \sin(a_{k+1} - a_k)$.Thus,the general term is $\frac{\sin(a_{k+1} - a_k)}{\sin a_k \sin a_{k+1}} = \frac{\sin a_{k+1} \cos a_k - \cos a_{k+1} \sin a_k}{\sin a_k \sin a_{k+1}} = \cot a_k - \cot a_{k+1}$.Substituting this into the sum:$S = (\cot a_1 - \cot a_2) + (\cot a_2 - \cot a_3) + \dots + (\cot a_{n-1} - \cot a_n)$.This is a telescoping series,so all intermediate terms cancel out:$S = \cot a_1 - \cot a_n$.

If $a_1, a_2, \dots, a_n$ are in $A.P.$ with common difference $d$,then the sum of the following series is $\sin d (\csc a_1 \csc a_2 + \csc a_2 \csc a_3 + \dots + \csc a_{n-1} \csc a_n)$

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