JEE MATHEMATICS MCQ: SEQUENCE OF SERIES

1 Hard

1 of 1

Analyzing +4 | -1

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

are in AP with common difference

d \ne 0

, then

\left( {\sin d} \right)\left[ {\sec {a_1}\sec {a_2} + \sec {a_2}\sec {a_3} + ... + \sec {a_{n - 1}}\sec {a_n}} \right]

is equal to

View answer and explanation

Correct option: C

\tan {a_n} - \tan {a_1}

Option C (

\tan {a_n} - \tan {a_1}

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

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