Let $\mathrm{a}_{1}, \mathrm{a}_{2}, \mathrm{a}_{3} \ldots \mathrm{a}_{\mathrm{n}}$ be n positive consecutive terms of an arithmetic progression. If $\mathrm{d}>0$ is its common difference, then \[\lim _{n \rightarrow \infty} \sqrt{\frac{d}{n}}\left(\frac{1}{\sqrt{a_{1}}+\sqrt{a_{2}}}+\frac{1}{\sqrt{a_{2}}+\sqrt{a_{3}}}+\ldots \ldots \ldots+\frac{1}{\sqrt{a_{n-1}}+\sqrt{a_{n}}}\right)\]

(C) $\frac{1}{\sqrt{\mathrm{~d}}}$

JEE Main · 2023 · 06 Apr 2023 · Shift 1 · Limits

Let

a_{1}, a_{2}, a_{3} \dots a_{n}

be n positive consecutive terms of an arithmetic progression. If

d > 0

is its common difference, then

n \to \infty lim \frac{d}{n} (\frac{1}{a _{1} + a _{2}} + \frac{1}{a _{2} + a _{3}} + \dots\dots\dots + \frac{1}{a _{n - 1} + a _{n}})

Answer: (A)

Rationalize each term in the sum:

\frac{1}{a _{k} + a _{k + 1}} = \frac{a _{k + 1} - a _{k}}{a _{k + 1} - a _{k}} = \frac{a _{k + 1} - a _{k}}{d}

The sum telescopes:

k = 1 \sum n - 1 \frac{1}{a _{k} + a _{k + 1}} = \frac{a _{n} - a _{1}}{d}

Since

a_{n} = a_{1} + (n - 1) d

a_{n} - a_{1} = a_{1} + (n - 1) d - a_{1}

For large

n

a_{n} \approx (n - 1) d

, so:

n \to \infty lim \frac{d}{n} \cdot \frac{a _{n} - a _{1}}{d} = n \to \infty lim \frac{d}{n} \cdot \frac{n d}{d} = n \to \infty lim \frac{d ^{2}}{d} = 1

JEE Main · 2023 · 06 Apr 2023 · Shift 1 · Single correct option