Let $$a_1, a_2, a_3, \ldots, a_n$$ be n positive consecutive terms of an arithmetic progression. If $$d > 0$$ is its common difference, then $$\lim_{n \to \infty} \sqrt{\dfrac{d}{n}}\dfrac{1}{\sqrt{a_1}+\sqrt{a_2}} + \dfrac{1}{\sqrt{a_2}+\sqrt{a_3}} + \ldots + \dfrac{1}{\sqrt{a_{n-1}}+\sqrt{a_n}}$$ is

Rationalizing each term in the sum:

$$ \frac{1}{\sqrt{a_k} + \sqrt{a_{k+1}}} = \frac{\sqrt{a_{k+1}} - \sqrt{a_k}}{a_{k+1} - a_k} = \frac{\sqrt{a_{k+1}} - \sqrt{a_k}}{d} $$

The sum telescopes:

$$ \sum_{k=1}^{n-1} \frac{1}{\sqrt{a_k} + \sqrt{a_{k+1}}} = \frac{\sqrt{a_n} - \sqrt{a_1}}{d} $$

The given expression becomes:

$$ \sqrt{\frac{d}{n}} \cdot \frac{\sqrt{a_n} - \sqrt{a_1}}{d} = \frac{\sqrt{a_n} - \sqrt{a_1}}{\sqrt{dn}} $$

Since $$a_n = a_1 + (n-1)d$$, as $$n \to \infty$$:

$$ \sqrt{a_n} = \sqrt{a_1 + (n-1)d} \approx \sqrt{nd} $$

Therefore:

$$ \lim_{n \to \infty} \frac{\sqrt{nd} - \sqrt{a_1}}{\sqrt{dn}} = \lim_{n \to \infty} \left(1 - \frac{\sqrt{a_1}}{\sqrt{dn}}\right) = 1 $$

The correct answer is 1.