Let a _ 1 , a _ 2 , a _ 3 , . . . . , a _ n be n positive...

Given,

${a}_{1}, {a}_{2}, {a}_{3}, . . . . , {a}_{n}$ are terms of an $A . P$,

So common difference will be,

$d = {a}_{2} - {a}_{1} = {a}_{3} - {a}_{2} = . . . . . . . . = {a}_{n} - {a}_{n - 1}$

Now solving,

$\lim_{n \to \infty}\sqrt{\dfrac{d}{n}}\left(\dfrac{1}{\sqrt{{a}_{1}} + \sqrt{{a}_{2}}} + \dfrac{1}{\sqrt{{a}_{2}} + \sqrt{{a}_{3}}} + \ldots + \dfrac{1}{\sqrt{{a}_{n - 1}} + \sqrt{{a}_{n}}}\right)$

$= \lim_{n \to \infty}\sqrt{\dfrac{d}{n}}\left(\dfrac{\sqrt{{a}_{2}} - \sqrt{{a}_{1}}}{{a}_{2} - {a}_{1}} + \dfrac{\sqrt{{a}_{3}} - \sqrt{{a}_{2}}}{{a}_{3} - {a}_{2}} + \ldots + \dfrac{\sqrt{{a}_{n}} - \sqrt{{a}_{n - 1}}}{{a}_{n} - {a}_{n - 1}}\right)$

$= \lim_{n \to \infty}\sqrt{\dfrac{d}{n}} \times \dfrac{1}{d}\left(\sqrt{{a}_{n}} - \sqrt{{a}_{1}}\right)$

Now using the formula ${a}_{n} = {a}_{1} + \left(n - 1\right)d$ we get,

$= \lim_{n \to \infty}\dfrac{1}{\sqrt{d}}\left(\dfrac{\sqrt{{a}_{1} + ( n - 1 ) d} - \sqrt{{a}_{1}}}{\sqrt{n}}\right)$

$= \lim_{n \to \infty}\dfrac{1}{\sqrt{d}}\left(\dfrac{\sqrt{n}\left(\sqrt{\dfrac{{a}_{1}}{n} + d - \dfrac{d}{n}} - \sqrt{\dfrac{{a}_{1}}{n}}\right)}{\sqrt{n}}\right)$

$= \dfrac{1}{\sqrt{d}}\left(\left(\sqrt{0 + d - 0} - \sqrt{0}\right)\right)$

$= \dfrac{1}{\sqrt{d}} \times \sqrt{d} = 1$