JEE MATHEMATICS MCQ: SEQUENCE OF SERIES

1 Easy

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Analyzing +4 | -1

Let the sequence

a_1, a_2, a_3, \ldots, a_n, \ldots

form an A.P. Then,

a_1^2-a_2^2+a_3^2-a_4^2+\ldots+a_{2 n-1}^2-a_{2 n}^2

is equal to

View answer and explanation

Correct option: A

\frac{n}{2 n-1}\left(a_1^2-a_{2 n}^2\right)

Option A (

\frac{n}{2 n-1}\left(a_1^2-a_{2 n}^2\right)

) is correct because it is derived by simplifying the given expression using the properties of an arithmetic progression.

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