Let a_1, a_2, a_3, \ldots, a_n be n positive consecutive terms of an arithmetic progression. If d \gt 0 is its common difference, then \lim_n arrow \infty \sqrt\fracdn(\frac1\sqrta_1+\sqrta_2+\frac1\sqrta_2+\sqrta_3+\ldots \ldots \ldots+\frac1\sqrta_n-1+\sqrta_n) is

Limits Continuity and Differentiability PYQ — JEE Main 2023 · 6 Apr · Shift 1 · Q32 with Solution

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Correct answer: B

Write the AP terms explicitly

Let the arithmetic progression be $a_k=a_1+(k-1)d, \qquad d>0.$ We need to evaluate $L_n=\sqrt{\frac{d}{n}}\sum_{k=1}^{n-1}\frac{1}{\sqrt{a_k}+\sqrt{a_{k+1}}}.$

Use the identity involving consecutive AP terms

Since $a_{k+1}-a_k=d,$ we use $(\sqrt{a_{k+1}}-\sqrt{a_k})(\sqrt{a_{k+1}}+\sqrt{a_k})=a_{k+1}-a_k=d.$ Hence, $\frac{1}{\sqrt{a_k}+\sqrt{a_{k+1}}}=\frac{\sqrt{a_{k+1}}-\sqrt{a_k}}{d}.$

Therefore,

=\frac{1}{d}\sum_{k=1}^{n-1}(\sqrt{a_{k+1}}-\sqrt{a_k}).$$ 3. **Observe the telescoping sum** The sum telescopes: $$\sum_{k=1}^{n-1}(\sqrt{a_{k+1}}-\sqrt{a_k})=\sqrt{a_n}-\sqrt{a_1}.$$ So, $$L_n=\sqrt{\frac{d}{n}}\cdot \frac{\sqrt{a_n}-\sqrt{a_1}}{d} =\frac{\sqrt{a_n}-\sqrt{a_1}}{\sqrt{dn}}.$$ 4. **Express $a_n$ in terms of $n$** Since $$a_n=a_1+(n-1)d,$$ we get $$L_n=\frac{\sqrt{a_1+(n-1)d}-\sqrt{a_1}}{\sqrt{dn}}.$$ 5. **Take the limit as $n\to\infty$** Now, $$\frac{\sqrt{a_1+(n-1)d}}{\sqrt{dn}} =\sqrt{\frac{a_1+(n-1)d}{dn}} =\sqrt{\frac{a_1}{dn}+\frac{n-1}{n}}\to 1.$$ Also, $$\frac{\sqrt{a_1}}{\sqrt{dn}}\to 0.$$ Thus, $$\lim_{n\to\infty}L_n=1-0=1.$$ 6. **Check options** - A: $\frac{1}{\sqrt d}$ — incorrect - B: $1$ — correct - C: $0$ — incorrect - D: $\sqrt d$ — incorrect Therefore, the required limit is $$\boxed{1}.$$

More from Limits Continuity and Differentiability

View written solutionFree

Correct answer: B

Write the AP terms explicitly

Let the arithmetic progression be $a_k=a_1+(k-1)d, \qquad d>0.$ We need to evaluate $L_n=\sqrt{\frac{d}{n}}\sum_{k=1}^{n-1}\frac{1}{\sqrt{a_k}+\sqrt{a_{k+1}}}.$

Use the identity involving consecutive AP terms

Since $a_{k+1}-a_k=d,$ we use $(\sqrt{a_{k+1}}-\sqrt{a_k})(\sqrt{a_{k+1}}+\sqrt{a_k})=a_{k+1}-a_k=d.$ Hence, $\frac{1}{\sqrt{a_k}+\sqrt{a_{k+1}}}=\frac{\sqrt{a_{k+1}}-\sqrt{a_k}}{d}.$

Therefore,

=\frac{1}{d}\sum_{k=1}^{n-1}(\sqrt{a_{k+1}}-\sqrt{a_k}).$$ 3. **Observe the telescoping sum** The sum telescopes: $$\sum_{k=1}^{n-1}(\sqrt{a_{k+1}}-\sqrt{a_k})=\sqrt{a_n}-\sqrt{a_1}.$$ So, $$L_n=\sqrt{\frac{d}{n}}\cdot \frac{\sqrt{a_n}-\sqrt{a_1}}{d} =\frac{\sqrt{a_n}-\sqrt{a_1}}{\sqrt{dn}}.$$ 4. **Express $a_n$ in terms of $n$** Since $$a_n=a_1+(n-1)d,$$ we get $$L_n=\frac{\sqrt{a_1+(n-1)d}-\sqrt{a_1}}{\sqrt{dn}}.$$ 5. **Take the limit as $n\to\infty$** Now, $$\frac{\sqrt{a_1+(n-1)d}}{\sqrt{dn}} =\sqrt{\frac{a_1+(n-1)d}{dn}} =\sqrt{\frac{a_1}{dn}+\frac{n-1}{n}}\to 1.$$ Also, $$\frac{\sqrt{a_1}}{\sqrt{dn}}\to 0.$$ Thus, $$\lim_{n\to\infty}L_n=1-0=1.$$ 6. **Check options** - A: $\frac{1}{\sqrt d}$ — incorrect - B: $1$ — correct - C: $0$ — incorrect - D: $\sqrt d$ — incorrect Therefore, the required limit is $$\boxed{1}.$$

Limits Continuity and Differentiability question

Limits Continuity and Differentiability question

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Limits Continuity and Differentiability question

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