JEE MAIN - Mathematics (2023 - 6th April Morning Shift - No. 10)
Let $$a_{1}, a_{2}, a_{3}, \ldots, a_{\mathrm{n}}$$ be $$\mathrm{n}$$ positive consecutive terms of an arithmetic progression. If $$\mathrm{d} > 0$$ is its common difference, then
$$\lim_\limits{n \rightarrow \infty} \sqrt{\frac{d}{n}}\left(\frac{1}{\sqrt{a_{1}}+\sqrt{a_{2}}}+\frac{1}{\sqrt{a_{2}}+\sqrt{a_{3}}}+\ldots \ldots \ldots+\frac{1}{\sqrt{a_{n-1}}+\sqrt{a_{n}}}\right)$$ is
$$\frac{1}{\sqrt{d}}$$
1
0
$$\sqrt{d}$$
Explanation
$$\lim_\limits{n \rightarrow \infty} \sqrt{\frac{d}{n}}\left(\frac{1}{\sqrt{a_{1}}+\sqrt{a_{2}}}+\frac{1}{\sqrt{a_{2}}+\sqrt{a_{3}}}+\ldots \ldots \ldots+\frac{1}{\sqrt{a_{n-1}}+\sqrt{a_{n}}}\right)$$
Now,
$\begin{aligned} & \frac{1}{\sqrt{a_1}+\sqrt{a_2}}+\frac{1}{\sqrt{a_2}+\sqrt{a_3}}+\ldots+\frac{1}{\sqrt{a_{n-1}}+\sqrt{a_n}} \\\\ = & \frac{\sqrt{a_2}-\sqrt{a_1}}{a_2-a_1}+\frac{\sqrt{a_3}-\sqrt{a_2}}{a_3-a_2}+\ldots+\frac{\sqrt{a_n}-\sqrt{a_{n-1}}}{a_n-a_{n-1}} \\\\ = & \frac{\sqrt{a_2}-\sqrt{a_1}+\sqrt{a_3}-\sqrt{a_2}+. .+\sqrt{a_n}-\sqrt{a_{n-1}}}{d}\end{aligned}$
$\left(\because a_2-a_1=a_3-a_2=\ldots a_n-a_{n-1}=d\right)$
$\begin{aligned} & =\frac{\sqrt{a_n}-\sqrt{a_1}}{d} \\\\ & =\frac{\sqrt{a_1+(n-1) d}-\sqrt{a_1}}{d}\end{aligned}$
$\therefore \lim\limits_{n \rightarrow \infty} \sqrt{\frac{d}{n}}\left(\frac{\sqrt{a_1+(n-1) d}-\sqrt{a_1}}{d}\right)$
$\begin{aligned} & =\lim _{n \rightarrow \infty}\left[\frac{1}{\sqrt{d}}\left(\frac{\sqrt{a_1+(n-1) d}-\sqrt{a_1}}{\sqrt{n}}\right)\right] \\\\ & =\lim _{n \rightarrow \infty}\left[\frac{1}{\sqrt{d}}\left(\sqrt{\frac{a_1}{n}+\left(d-\frac{d}{n}\right)}-\sqrt{\frac{a_1}{n}}\right)\right] \\\\ & =\frac{1}{\sqrt{d}}(\sqrt{0+d-0}-\sqrt{0})=\frac{\sqrt{d}}{\sqrt{d}}=1\end{aligned}$
Now,
$\begin{aligned} & \frac{1}{\sqrt{a_1}+\sqrt{a_2}}+\frac{1}{\sqrt{a_2}+\sqrt{a_3}}+\ldots+\frac{1}{\sqrt{a_{n-1}}+\sqrt{a_n}} \\\\ = & \frac{\sqrt{a_2}-\sqrt{a_1}}{a_2-a_1}+\frac{\sqrt{a_3}-\sqrt{a_2}}{a_3-a_2}+\ldots+\frac{\sqrt{a_n}-\sqrt{a_{n-1}}}{a_n-a_{n-1}} \\\\ = & \frac{\sqrt{a_2}-\sqrt{a_1}+\sqrt{a_3}-\sqrt{a_2}+. .+\sqrt{a_n}-\sqrt{a_{n-1}}}{d}\end{aligned}$
$\left(\because a_2-a_1=a_3-a_2=\ldots a_n-a_{n-1}=d\right)$
$\begin{aligned} & =\frac{\sqrt{a_n}-\sqrt{a_1}}{d} \\\\ & =\frac{\sqrt{a_1+(n-1) d}-\sqrt{a_1}}{d}\end{aligned}$
$\therefore \lim\limits_{n \rightarrow \infty} \sqrt{\frac{d}{n}}\left(\frac{\sqrt{a_1+(n-1) d}-\sqrt{a_1}}{d}\right)$
$\begin{aligned} & =\lim _{n \rightarrow \infty}\left[\frac{1}{\sqrt{d}}\left(\frac{\sqrt{a_1+(n-1) d}-\sqrt{a_1}}{\sqrt{n}}\right)\right] \\\\ & =\lim _{n \rightarrow \infty}\left[\frac{1}{\sqrt{d}}\left(\sqrt{\frac{a_1}{n}+\left(d-\frac{d}{n}\right)}-\sqrt{\frac{a_1}{n}}\right)\right] \\\\ & =\frac{1}{\sqrt{d}}(\sqrt{0+d-0}-\sqrt{0})=\frac{\sqrt{d}}{\sqrt{d}}=1\end{aligned}$
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