JEE MATHEMATICS MCQ: SEQUENCE OF SERIES

1 Easy

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Analyzing +4 | -1

Let

a_{1}, a_{2}, a_{3}, \ldots ., a_{n}

n

positive consecutive terms of an arithmetic progression. If

d>0

is its common difference, then

\lim \limits_{{n \rightarrow \infty}} \sqrt{\frac{d}{n}}\left(\begin{array}{c} \frac{1}{\sqrt{a_{1}}+\sqrt{a_{2}}}+\frac{1}{\sqrt{a_{2}}+\sqrt{a_{3}}}+ \ldots+\frac{1}{\sqrt{a_{n-1}}+\sqrt{a_{n}}} \end{array}\right)

View answer and explanation

Correct option: C

Option C (1) is correct because the limit simplifies to 1 after rationalizing the denominator and applying limit properties.

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