JEE Main 2023 — Mathematics Algebra

Given,

${a}_{1},{a}_{2},{a}_{3},....,{a}_{n}$a1​,a2​,a3​,....,an​ are terms of an $A.P$A.P,

So common difference will be,

$$d={a}_{2}-{a}_{1}={a}_{3}-{a}_{2}=........={a}_{n}-{a}_{n-1}$$d=a2​−a1​=a3​−a2​=........=an​−an−1​

Now solving,

$$\underset{n\rightarrow \infty }{\mathrm{lim}}\sqrt{\frac{d}{n}}(\frac{1}{\sqrt{{a}_{1}}+\sqrt{{a}_{2}}}+\frac{1}{\sqrt{{a}_{2}}+\sqrt{{a}_{3}}}+\ldots +\frac{1}{\sqrt{{a}_{n-1}}+\sqrt{{a}_{n}}})$$n→∞lim​nd​​(a1​​+a2​​1​+a2​​+a3​​1​+…+an−1​​+an​​1​)

$$=\underset{n\rightarrow \infty }{\mathrm{lim}}\sqrt{\frac{d}{n}}(\frac{\sqrt{{a}_{2}}-\sqrt{{a}_{1}}}{{a}_{2}-{a}_{1}}+\frac{\sqrt{{a}_{3}}-\sqrt{{a}_{2}}}{{a}_{3}-{a}_{2}}+\ldots +\frac{\sqrt{{a}_{n}}-\sqrt{{a}_{n-1}}}{{a}_{n}-{a}_{n-1}})$$=n→∞lim​nd​​(a2​−a1​a2​​−a1​​​+a3​−a2​a3​​−a2​​​+…+an​−an−1​an​​−an−1​​​)

$$=\underset{n\rightarrow \infty }{\mathrm{lim}}\sqrt{\frac{d}{n}}\times \frac{1}{d}(\sqrt{{a}_{n}}-\sqrt{{a}_{1}})$$=n→∞lim​nd​​×d1​(an​​−a1​​)

Now using the formula ${a}_{n}={a}_{1}+(n-1)d$an​=a1​+(n−1)d we get,

$$=\underset{n\rightarrow \infty }{\mathrm{lim}}\frac{1}{\sqrt{d}}(\frac{\sqrt{{a}_{1}+(n-1)d}-\sqrt{{a}_{1}}}{\sqrt{n}})$$=n→∞lim​d​1​(n​a1​+(n−1)d​−a1​​​)

$$=\underset{n\rightarrow \infty }{\mathrm{lim}}\frac{1}{\sqrt{d}}(\frac{\sqrt{n}(\sqrt{\frac{{a}_{1}}{n}+d-\frac{d}{n}}-\sqrt{\frac{{a}_{1}}{n}})}{\sqrt{n}})$$=n→∞lim​d​1​(n​n​(na1​​+d−nd​​−na1​​​)​)

$$=\frac{1}{\sqrt{d}}((\sqrt{0+d-0}-\sqrt{0}))$$=d​1​((0+d−0​−0​))

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