JEE MATHEMATICS MCQ: SEQUENCE OF SERIES

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Analyzing +4 | -1

a_1, a_2, a_3, \ldots ., a_{2 n+1}

are in A.P., then

\frac{a_{2 n+1}-a_1}{a_{2 n+1}+a_1}+\frac{a_{2 n}-a_2}{a_{2 n}+a_2}+\ldots .+\frac{a_{n+2}-a_n}{a_{n+2}+a_n}

is equal to

View answer and explanation

Correct option: A

\frac{n(n+1)}{2} \cdot \frac{a_2-a_1}{a_{n+1}}

Option A (

\frac{n(n+1)}{2} \cdot \frac{a_2-a_1}{a_{n+1}}

) is correct because the given expression simplifies to this form using the properties of an arithmetic progression.

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