Let $a_1, a_2, a_3, \dots, a_n$ be $n$ positive consecutive terms of an arithmetic progression. If $d > 0$ is its common difference, then $\lim_{n \to \infty} \sqrt{\frac{d}{n}} \left( \frac{1}{\sqrt{a_1} + \sqrt{a_2}} + \frac{1}{\sqrt{a_2} + \sqrt{a_3}} + \dots + \frac{1}{\sqrt{a_{n-1}} + \sqrt{a_n}} \right)$ is Options: (A) $\frac{1}{\sqrt{d}}$ (B) $\sqrt{d}$ (C) 1 (D) 2

### Analyzing the Setup The problem asks us to evaluate the limit: $$L = \lim_{n \to \infty} \sqrt{\frac{d}{n}} \sum_{k=1}^{n-1} \frac{1}{\sqrt{a_k} + \sqrt{a_{k+1}}}$$ Here, $\{a_k\}$ is an Arithmetic Progression (A.P.) with a common difference $d > 0$. The presence of square roots in the denominator is a classic cue to apply the method of rationalization. ### The Power of Conjugates Consider a single term from the summation: $$\frac{1}{\sqrt{a_k} + \sqrt{a_{k+1}}}$$ To rationalize, we multiply the numerator and the denominator by the conjugate, $\sqrt{a_{k+1}} - \sqrt{a_k}$. Using the identity $(A+B)(A-B) = A^2 - B^2$, the denominator becomes: $$(\sqrt{a_{k+1}})^2 - (\sqrt{a_k})^2 = a_{k+1} - a_k = d$$ Thus, the term simplifies to: $$\frac{\sqrt{a_{k+1}} - \sqrt{a_k}}{d}$$ ### The Telescoping Dance Now, we substitute this simplified form back into the summation: $$\frac{1}{d} \sum_{k=1}^{n-1} (\sqrt{a_{k+1}} - \sqrt{a_k})$$ Expanding this sum reveals the **Telescoping Effect**: $$\frac{1}{d} [(\sqrt{a_2} - \sqrt{a_1}) + (\sqrt{a_3} - \sqrt{a_2}) + \dots + (\sqrt{a_n} - \sqrt{a_{n-1}})]$$ All intermediate terms cancel out, leaving only the first and last components: $$\frac{1}{d} (\sqrt{a_n} - \sqrt{a_1})$$ ### The Final Limit Substitute this result into the original limit expression: $$L = \lim_{n \to \infty} \sqrt{\frac{d}{n}} \cdot \frac{1}{d} (\sqrt{a_n} - \sqrt{a_1}) = \lim_{n \to \infty} \frac{1}{\sqrt{nd}} (\sqrt{a_n} - \sqrt{a_1})$$ Given that $a_n = a_1 + (n-1)d$, we substitute for $a_n$: $$L = \lim_{n \to \infty} \frac{\sqrt{a_1 + (n-1)d} - \sqrt{a_1}}{\sqrt{nd}}$$ To evaluate this as $n \to \infty$, factor $\sqrt{n}$ out of the numerator: $$L = \lim_{n \to \infty} \frac{\sqrt{n} \left( \sqrt{\frac{a_1}{n} + d - \frac{d}{n}} - \sqrt{\frac{a_1}{n}} \right)}{\sqrt{n} \cdot \sqrt{d}}$$ As $n \to \infty$, the terms $\frac{a_1}{n}$ and $\frac{d}{n}$ approach $0$. This leaves us with: $$L = \frac{\sqrt{d}}{\sqrt{d}} = 1$$ The final value of the limit is **1**.

JEE Main 2023 (06 Apr Shift 1)

LEVELJEE Main

Animated Solution for Mathematics - Sequence and Series: Let $a_{1}, a_{2}, a_{3}, \dots, a_{n}$ be $n$ positive consecutive terms of an arithmetic progression. If $d > 0$ is its common difference, then $lim_{n \to \infty} \frac{d}{n} (\frac{1}{a _{1} + a _{2}} + \frac{1}{a _{2} + a _{3}} + \dots + \frac{1}{a _{n - 1} + a _{n}})$ is

Q: How do I know for sure that it's a telescoping series?

Look at the structure: $f(k+1) - f(k)$. If your sum looks like $\sum [f(k+1) - f(k)]$, it is telescoping. The middle terms will always vanish, leaving only the boundary values. It is one of the most elegant patterns in mathematics.

Q: Why does the $a_1$ term disappear in the limit?

As $n$ approaches infinity, the term $\sqrt{a_1}$ becomes insignificant compared to the growing $\sqrt{n}$ term. Mathematically, $\frac{\sqrt{a_1}}{\sqrt{n}} \to 0$ as $n \to \infty$. We are essentially looking at the dominant behavior of the expression.

Select Answer:

Progressive Hints

Hint 1

The denominator is a sum of square roots. In the world of JEE Advanced, this is a massive red flag screaming for rationalization. Don't try to sum them directly; transform the expression into a difference of square roots using the conjugate.

Hint 2

Once you rationalize, you will see a pattern emerge. Write out the first few terms of the summation. You will notice that the second part of the first term cancels the first part of the second term. This is the 'Telescoping Effect'—a beautiful, rhythmic collapse of the series.

Hint 3

After the telescoping, you are left with

\frac{1}{d} (a_{n} - a_{1})

. Now, substitute the general term of the A.P.,

a_{n} = a_{1} + (n - 1) d

. To evaluate the limit as

n \to \infty

, factor out

n

from the numerator and denominator. The terms involving

\frac{1}{n}

will vanish, leaving you with the final answer.

Visualized Solution

Understanding the Arithmetic Progression

Given $a_{1}, a_{2}, \dots, a_{n}$ are in an Arithmetic Progression.
Common difference $d = a_{2} - a_{1} = a_{3} - a_{2} = \dots > 0$ .
Objective: Evaluate $lim_{n \to \infty} \frac{d}{n} \sum_{k = 1}^{n - 1} \frac{1}{a _{k} + a _{k + 1}}$ .

Rationalizing the Terms

To simplify $\frac{1}{a _{k} + a _{k + 1}}$ , we rationalize the denominator.
Multiply numerator and denominator by the conjugate: $(a_{k + 1} - a_{k})$ .

Simplifying the General Term

The denominator becomes $a_{k + 1} - a_{k}$ .
Since the terms are in A.P., $a_{k + 1} - a_{k} = d$ .
The term simplifies to $\frac{1}{d} (a_{k + 1} - a_{k})$ .

Expanding the Summation

Let's write out the terms of the sum explicitly.
$S = \frac{1}{d} [(a_{2} - a_{1}) + (a_{3} - a_{2}) + \dots + (a_{n} - a_{n - 1})]$

The Telescoping Effect

Notice that $a_{2}$ from the first term cancels with $- a_{2}$ from the second term.

Canceling Intermediate Terms

Similarly, $a_{3}$ and all intermediate terms up to $- a_{n - 1}$ will cancel out.

The Simplified Sum

Only the last positive term and the first negative term survive.
$S = \frac{1}{d} (a_{n} - a_{1})$

Updating the Limit Expression

Substitute the simplified sum back into the original limit expression.
$L = lim_{n \to \infty} \frac{d}{n} \cdot \frac{1}{d} (a_{n} - a_{1})$

Using the A.P. General Term

Substitute $a_{n} = a_{1} + (n - 1) d$ .
Simplify the constants: $\frac{d}{d} = \frac{1}{d}$ .
$L = lim_{n \to \infty} \frac{1}{n d} (a_{1} + (n - 1) d - a_{1})$

Factoring Out the Dominant Term

To evaluate the limit as $n \to \infty$ , factor out $n$ from the numerator.
$L = lim_{n \to \infty} \frac{n}{n d} (\frac{a _{1}}{n} + d - \frac{d}{n} - \frac{a _{1}}{n})$

Evaluating the Limit

As $n \to \infty$ , terms like $\frac{a _{1}}{n}$ and $\frac{d}{n}$ approach $0$ .
The expression simplifies to $\frac{d}{d} = 1$ .

00:00 / 00:00

The Sigma Insight: Arithmetic Progression (A.P.)

Solution Diagram

Analyzing the Setup

The problem asks us to evaluate the limit:

L = n \to \infty lim \frac{d}{n} k = 1 \sum n - 1 \frac{1}{a _{k} + a _{k + 1}}

Here,

{a_{k}}

is an Arithmetic Progression (A.P.) with a common difference

d > 0

. The presence of square roots in the denominator is a classic cue to apply the method of rationalization.

The Power of Conjugates

Consider a single term from the summation:

\frac{1}{a _{k} + a _{k + 1}}

To rationalize, we multiply the numerator and the denominator by the conjugate,

a_{k + 1} - a_{k}

. Using the identity

(A + B) (A - B) = A^{2} - B^{2}

, the denominator becomes:

(a_{k + 1})^{2} - (a_{k})^{2} = a_{k + 1} - a_{k} = d

Thus, the term simplifies to:

\frac{a _{k + 1} - a _{k}}{d}

The Telescoping Dance

Now, we substitute this simplified form back into the summation:

\frac{1}{d} k = 1 \sum n - 1 (a_{k + 1} - a_{k})

Expanding this sum reveals the Telescoping Effect:

\frac{1}{d} [(a_{2} - a_{1}) + (a_{3} - a_{2}) + \dots + (a_{n} - a_{n - 1})]

All intermediate terms cancel out, leaving only the first and last components:

\frac{1}{d} (a_{n} - a_{1})

The Final Limit

Substitute this result into the original limit expression:

L = n \to \infty lim \frac{d}{n} \cdot \frac{1}{d} (a_{n} - a_{1}) = n \to \infty lim \frac{1}{n d} (a_{n} - a_{1})

Given that

a_{n} = a_{1} + (n - 1) d

, we substitute for

a_{n}

L = n \to \infty lim \frac{a _{1} + ( n - 1 ) d - a _{1}}{n d}

To evaluate this as

n \to \infty

, factor

n

out of the numerator:

L = n \to \infty lim \frac{n ( \frac{a _{1}}{n} + d - \frac{d}{n} - \frac{a _{1}}{n} )}{n \cdot d}

n \to \infty

, the terms

\frac{a _{1}}{n}

and

\frac{d}{n}

approach

0

. This leaves us with:

L = \frac{d}{d} = 1

The final value of the limit is 1.

Common Traps & FAQs

Why do we rationalize? Can't we just use an approximation for large

n

Rationalization is the only way to get an exact, clean result. Approximations are risky in JEE Advanced because they often hide subtle dependencies on

d

n

that only reveal themselves after the algebra is simplified. Always seek the exact algebraic path first.

How do I know for sure that it's a telescoping series?

Look at the structure:

f (k + 1) - f (k)

. If your sum looks like

\sum [f (k + 1) - f (k)]

, it is telescoping. The middle terms will always vanish, leaving only the boundary values. It is one of the most elegant patterns in mathematics.

Why does the

a_{1}

term disappear in the limit?

n

approaches infinity, the term

a_{1}

becomes insignificant compared to the growing

n

term. Mathematically,

\frac{a _{1}}{n} \to 0

n \to \infty

. We are essentially looking at the dominant behavior of the expression.

Select Answer:

Visualized Solution

Understanding the Arithmetic Progression

Rationalizing the Terms

Simplifying the General Term

Expanding the Summation

The Telescoping Effect

Canceling Intermediate Terms

The Simplified Sum

Updating the Limit Expression

Using the A.P. General Term

Factoring Out the Dominant Term

Evaluating the Limit

The Sigma Insight: Arithmetic Progression (A.P.)

Analyzing the Setup

The Power of Conjugates

The Telescoping Dance

The Final Limit

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Comprehension Passage

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Select Answer:

Visualized Solution

Understanding the Arithmetic Progression

Rationalizing the Terms

Simplifying the General Term

Expanding the Summation

The Telescoping Effect

Canceling Intermediate Terms

The Simplified Sum

Updating the Limit Expression

Using the A.P. General Term

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The Sigma Insight: Arithmetic Progression (A.P.)

Analyzing the Setup

The Power of Conjugates

The Telescoping Dance

The Final Limit

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