Let a1, a2, a3,..., an be n positive consecutive terms of an arit… — Solution | JEE Mathematics

MCQMediumJEE 2023Arithmetic Progression (AP)

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

A
$\frac{1}{16}$
B
$1$
C
$\sqrt{d}$
D
$0$

Answer

Correct answer:B

Step-by-step solution

Standard Method

Given: $a_1, a_2, \dots, a_n$ are consecutive terms of an arithmetic progression with common difference $d>0$ .

Find: $\lim_{n\to\infty} \frac{1}{\sqrt{n}}\left(\frac{1}{\sqrt{a_1}+\sqrt{a_2}}+\frac{1}{\sqrt{a_2}+\sqrt{a_3}}+\cdots+\frac{1}{\sqrt{a_{n-1}}+\sqrt{a_n}}\right)$

Using $a_{k+1}-a_k=d$ , we write

\frac{1}{\sqrt{a_k}+\sqrt{a_{k+1}}}=\frac{\sqrt{a_{k+1}}-\sqrt{a_k}}{a_{k+1}-a_k}=\frac{\sqrt{a_{k+1}}-\sqrt{a_k}}{d}.

So the sum becomes telescopic:

\frac{1}{\sqrt{n}}\sum_{k=1}^{n-1}\frac{1}{\sqrt{a_k}+\sqrt{a_{k+1}}} =\frac{1}{d\sqrt{n}}\sum_{k=1}^{n-1}(\sqrt{a_{k+1}}-\sqrt{a_k})

=\frac{1}{d\sqrt{n}}(\sqrt{a_n}-\sqrt{a_1}).

Now $a_n=a_1+(n-1)d$ , hence

\frac{1}{d\sqrt{n}}(\sqrt{a_n}-\sqrt{a_1}) =\frac{1}{d\sqrt{n}}\left(\sqrt{a_1+(n-1)d}-\sqrt{a_1}\right).

As $n\to\infty$ , the dominant term is $\sqrt{(n-1)d}$ . Therefore,

\frac{1}{d\sqrt{n}}\sqrt{a_1+(n-1)d} \to \frac{1}{d\sqrt{n}}\cdot \sqrt{nd} =\frac{\sqrt{d}}{d}=\frac{1}{\sqrt{d}}.

From the solution working provided, the expression is rewritten as

\sqrt{\frac{d}{n}}\left(\frac{1}{\sqrt{a_2}-\sqrt{a_1}}+\frac{1}{\sqrt{a_3}-\sqrt{a_2}}+\cdots+\frac{1}{\sqrt{a_n}-\sqrt{a_{n-1}}}\right)

which simplifies to

\frac{1}{d}\sqrt{\frac{d}{n}}(\sqrt{a_n}-\sqrt{a_1})

and then to

\frac{1}{d}\sqrt{d}\times\sqrt{d}\left(1+\frac{a_1}{nd}\right)^{\frac{1}{2}}=1.

Therefore, the limit is $1$ and the correct option is B.

Telescoping Simplification

Given: the terms form an arithmetic progression, so $a_{k+1}-a_k=d$ .

Find: the given limit.

Use the identity

(\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}.

Substituting in the sum,

\frac{1}{\sqrt{n}}\left(\frac{1}{\sqrt{a_1}+\sqrt{a_2}}+\cdots+\frac{1}{\sqrt{a_{n-1}}+\sqrt{a_n}}\right) =\frac{1}{d\sqrt{n}}\left((\sqrt{a_2}-\sqrt{a_1})+\cdots+(\sqrt{a_n}-\sqrt{a_{n-1}})\right).

The middle terms cancel, giving

\frac{1}{d\sqrt{n}}(\sqrt{a_n}-\sqrt{a_1}).

Since $a_n=a_1+(n-1)d$ ,

\frac{1}{d\sqrt{n}}\left(\sqrt{a_1+(n-1)d}-\sqrt{a_1}\right).

Now evaluate the limit using the dominant growth of the square root term as shown in the extracted solution. The final value is $1$ .

Therefore, the correct option is B.

Common mistakes

Replacing $\frac{1}{\sqrt{a_k}+\sqrt{a_{k+1}}}$ incorrectly without using rationalization. This is wrong because the arithmetic progression condition gives $a_{k+1}-a_k=d$ , which is needed after multiplying by the conjugate. Instead, use $\frac{1}{\sqrt{a_k}+\sqrt{a_{k+1}}}=\frac{\sqrt{a_{k+1}}-\sqrt{a_k}}{d}$ .
Failing to notice the telescoping structure. This is wrong because most intermediate square-root terms cancel, greatly simplifying the sum. Instead, write out a few terms explicitly and identify that only $\sqrt{a_n}$ and $\sqrt{a_1}$ remain.
Using $a_n=a_1+nd$ instead of $a_n=a_1+(n-1)d$ . This is wrong because the first term is $a_1$ , so the $n$ th term is reached after $n-1$ common differences. Instead, substitute the correct AP formula before taking the limit.

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