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
$\lim_{n \to \infty} \sqrt{\frac{d}{n}} (\frac{1}{\sqrt{a_{1}} + \sqrt{a_{2}}} + \frac{1}{\sqrt{a_{2}} + \sqrt{a_{3}}} + \dots \dots . + \frac{1}{\sqrt{a_{n - 1}} + \sqrt{a_{n}}})$ is

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$\frac{1}{\sqrt{d}}$

$\sqrt{d}$

answer is B.

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Detailed Solution

$\lim_{n \to \infty} \sqrt{\frac{d}{n}} (\frac{1}{\sqrt{a_{1}} + \sqrt{a_{2}}} + \frac{1}{\sqrt{a_{1}} + \sqrt{a_{3}}} + \dots \dots . . + \frac{1}{\sqrt{a_{n - 1}} + \sqrt{a_{n}}})$

Rationalize the fraction

$\lim_{n \to \infty} \sqrt{\frac{d}{n}} (\frac{\sqrt{a_{2}} - \sqrt{a_{1}}}{a_{2} - a_{1}} + \frac{\sqrt{a_{3}} - \sqrt{a_{2}}}{a_{3} - a_{2}} + \dots \dots . . + \frac{\sqrt{a_{n}} - \sqrt{a_{n - 1}}}{a_{n} - a_{n - 1}})$

$a_{2} - a_{1} = a_{3} - a_{2} = d = a_{4} - a_{3} = \dots \dots . = a_{n} - a_{n - 1}$

$\lim_{n \to \infty} \sqrt{\frac{d}{n}} (\frac{\sqrt{a_{2}} - \sqrt{a_{1}}}{d} + \frac{\sqrt{a_{3}} - \sqrt{a_{2}}}{d} + \dots \dots + \frac{\sqrt{a_{n}} - \sqrt{a_{n - 1}}}{d}) = \lim_{n \to \infty} \sqrt{\frac{d}{n}} (\frac{\sqrt{a_{n}} - \sqrt{a_{1}}}{d})$

$= \lim_{n \to \infty} \frac{1}{\sqrt{d}} [\frac{\sqrt{a_{1} + (n - 1) d} - \sqrt{a_{1}}}{\sqrt{n}}] = \frac{1}{\sqrt{d}} \underset{n \to \infty}{Lt} [\sqrt{\frac{a_{1}}{n} + (1 - \frac{1}{n}) d} - \sqrt{\frac{a_{1}}{n}}] = \frac{1}{\sqrt{d}} \times \sqrt{d} = 1$