Let a_1,a_2,a_3, ,a_n be n positive consecutive terms of an arithmetic progression.

Question

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

$\underset{n	o \infty}{lim}\sqrt{\frac{d}{n}}(\frac{1}{\sqrt{{a}_{1}}+\sqrt{{a}_{2}}}+\frac{1}{\sqrt{{a}_{2}}+\sqrt{{a}_{3}}}+\ldots \ldots \ldots +\frac{1}{\sqrt{{a}_{n-1}}+\sqrt{{a}_{n}}})$ is

Accepted Answer

$\underset{n	o \infty}{lim}\sqrt{\frac{d}{n}}(\frac{1}{\sqrt{{a}_{1}}+\sqrt{{a}_{2}}}+\frac{1}{\sqrt{{a}_{2}}+\sqrt{{a}_{3}}}+\ldots \ldots \ldots +\frac{1}{\sqrt{{a}_{n-1}}+\sqrt{{a}_{n}}})$

Now,

$\begin{gathered} \frac{1}{\sqrt{{a}_{1}}+\sqrt{{a}_{2}}}+\frac{1}{\sqrt{{a}_{2}}+\sqrt{{a}_{3}}}+\ldots +\frac{1}{\sqrt{{a}_{n-1}}+\sqrt{{a}_{n}}} \ \ = \frac{\sqrt{{a}_{2}}-\sqrt{{a}_{1}}}{{a}_{2}-{a}_{1}}+\frac{\sqrt{{a}_{3}}-\sqrt{{a}_{2}}}{{a}_{3}-{a}_{2}}+\ldots +\frac{\sqrt{{a}_{n}}-\sqrt{{a}_{n-1}}}{{a}_{n}-{a}_{n-1}} \ \ = \frac{\sqrt{{a}_{2}}-\sqrt{{a}_{1}}+\sqrt{{a}_{3}}-\sqrt{{a}_{2}}+..+\sqrt{{a}_{n}}-\sqrt{{a}_{n-1}}}{d} \end{gathered}$

$(\because {a}_{2}-{a}_{1}={a}_{3}-{a}_{2}=\ldots {a}_{n}-{a}_{n-1}=d)$

$\begin{gathered} =\frac{\sqrt{{a}_{n}}-\sqrt{{a}_{1}}}{d} \ =\frac{\sqrt{{a}_{1}+(n-1)d}-\sqrt{{a}_{1}}}{d} \end{gathered}$

$	herefore \underset{n	o \infty}{lim}\sqrt{\frac{d}{n}}(\frac{\sqrt{{a}_{1}+(n-1)d}-\sqrt{{a}_{1}}}{d})$

$\begin{gathered} =\underset{n	o \infty}{lim}[\frac{1}{\sqrt{d}}(\frac{\sqrt{{a}_{1}+(n-1)d}-\sqrt{{a}_{1}}}{\sqrt{n}})] \ =\underset{n	o \infty}{lim}[[\frac{1}{\sqrt{d}}((\sqrt{\frac{{a}_{1}}{n}+(d-\frac{d}{n})}-\sqrt{\frac{{a}_{1}}{n}}))]] \ =\frac{1}{\sqrt{d}}(\sqrt{0+d-0}-\sqrt{0})=\frac{\sqrt{d}}{\sqrt{d}}=1 \end{gathered}$

Solutions · 1