Let a 1 , a 2 , a 3 , . . . . , a n be n positive consecutive terms of an arithmetic progression. If d >…

Given,

$a_1,a_2,a_3,....,a_n$ are terms of an $\mathrm{A}.\mathrm{P}$, 

So common difference will be, 

$d=a_2-a_1=a_3-a_2=........=a_n-a_{n-1}$

Now solving,

$\underset{n\to \infty }{\lim }\sqrt{\frac{d}{n}}\left(\frac{1}{\sqrt{a_1}+\sqrt{a_2}}+\frac{1}{\sqrt{a_2}+\sqrt{a_3}}+\dots +\frac{1}{\sqrt{a_{n-1}}+\sqrt{a_n}}\right)$

$=\underset{n\to \infty }{\lim }\sqrt{\frac{d}{n}}\left(\frac{\sqrt{a_2}-\sqrt{a_1}}{a_2-a_1}+\frac{\sqrt{a_3}-\sqrt{a_2}}{a_3-a_2}+\dots +\frac{\sqrt{a_n}-\sqrt{a_{n-1}}}{a_n-a_{n-1}}\right)$

$=\underset{n\to \infty }{\lim }\sqrt{\frac{d}{n}}\times \frac{1}{d}\left(\sqrt{a_n}-\sqrt{a_1}\right)$

Now using the formula $a_n=a_1+\left(n-1\right)d$ we get,

$=\underset{n\to \infty }{\lim }\frac{1}{\sqrt{d}}\left(\frac{\sqrt{a_1+(n-1)d}-\sqrt{a_1}}{\sqrt{n}}\right)$

$=\underset{n\to \infty }{\lim }\frac{1}{\sqrt{d}}\left(\frac{\sqrt{n}\left(\sqrt{{\displaystyle \frac{a_1}{n}}+d-{\displaystyle \frac{d}{n}}}-\sqrt{{\displaystyle \frac{a_1}{n}}}\right)}{\sqrt{n}}\right)$

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

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