If ${a_1},\;{a_2}, \ldots ,\;{a_n}$ are in AP with common difference $d$, then the sum of the series $\sin d\;\left( {{\rm{cosec}}\;{a_1}\;{\rm{cosec\;}}{a_2} + {\rm{cosec}}\;{a_2}\;{\rm{cosec}}\;{a_3} + ... + {\rm{cosec}}\;{a_{n - 1}}\;{\rm{cosec}}\;{a_n}} \right)$ is

${\rm{cosec}}\;{a_1} = {\rm{cosec}}\;{a_n}$

JEE MATHEMATICS MCQ: SEQUENCE OF SERIES

1 Easy

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Analyzing +4 | -1

{a_1},\;{a_2}, \ldots ,\;{a_n}

are in AP with common difference

d

, then the sum of the series

\sin d\;\left( {\mathrm{cosec}\;{a_1}\;\mathrm{cosec\;}{a_2} + \mathrm{cosec}\;{a_2}\;\mathrm{cosec}\;{a_3} + ... + \mathrm{cosec}\;{a_{n - 1}}\;\mathrm{cosec}\;{a_n}} \right)

View answer and explanation

Correct option: B

\cot {a_1} - \cot {a_n}

Option B (

\cot {a_1} - \cot {a_n}

) is correct because the series telescopes, leaving only the first and last terms after simplification.