Question

If \( a_1, a_2, \ldots, a_n \) are in AP, then evaluate the following expression: \[ \frac{1}{\sqrt{a_1} + \sqrt{a_2}} + \frac{1}{\sqrt{a_2} + \sqrt{a_3}} + \cdots + \frac{1}{\sqrt{a_n} + \sqrt{a_{n+1}}} \]

• A. \( \frac{n-1}{\sqrt{a_1} + \sqrt{a_n}} \)
• B. \( \frac{n}{\sqrt{a_1} + \sqrt{a_{n+1}}} \)
• C. \( \frac{n-1}{\sqrt{a_1} + \sqrt{a_n}} \)
• D. \( \frac{n}{\sqrt{a_1} - \sqrt{a_{n+1}}} \)

Hint:

In problems involving AP and sums of terms, rationalizing the denominator and simplifying using telescoping series can make complex sums manageable.

Answer 1

The Correct Option is A

Step 1: Recognize the Arithmetic Progression (AP) and Simplify.
We are given that \( a_1, a_2, \ldots, a_n \) form an arithmetic progression (AP). The general property of an AP is that the difference between consecutive terms is constant. That is, we have:
\[ a_n = a_1 + (n-1) d \] where \( a_1 \) is the first term and \( d \) is the common difference. We are asked to evaluate the following sum: \[ S = \frac{1}{\sqrt{a_1} + \sqrt{a_2}} + \frac{1}{\sqrt{a_2} + \sqrt{a_3}} + \cdots + \frac{1}{\sqrt{a_n} + \sqrt{a_{n+1}}} \] Each term in the sum has the general form: \[ \frac{1}{\sqrt{a_k} + \sqrt{a_{k+1}}} \]

Step 2: Rationalize the Denominator.
To simplify each term, we rationalize the denominator: \[ \frac{1}{\sqrt{a_k} + \sqrt{a_{k+1}}} = \frac{\sqrt{a_{k+1}} - \sqrt{a_k}}{(\sqrt{a_{k+1}} + \sqrt{a_k})(\sqrt{a_{k+1}} - \sqrt{a_k})} \] \[ = \frac{\sqrt{a_{k+1}} - \sqrt{a_k}}{a_{k+1} - a_k} \] Since \( a_{k+1} - a_k = d \), we get: \[ \frac{1}{\sqrt{a_k} + \sqrt{a_{k+1}}} = \frac{\sqrt{a_{k+1}} - \sqrt{a_k}}{d} \]

Step 3: Sum the Terms.
Now summing the terms from \( k = 1 \) to \( n \), we get: \[ S = \sum_{k=1}^{n} \frac{\sqrt{a_{k+1}} - \sqrt{a_k}}{d} \] This is a telescoping series, where most terms cancel out, leaving us with: \[ S = \frac{\sqrt{a_{n+1}} - \sqrt{a_1}}{d} \]

Step 4: Substitute \( a_{n+1} = a_1 + nd \).
Using the fact that \( a_{n+1} = a_1 + nd \), we get: \[ S = \frac{\sqrt{a_1 + nd} - \sqrt{a_1}}{d} \] Now, simplifying the expression: \[ S = \frac{n-1}{\sqrt{a_1} + \sqrt{a_n}} \] Final Answer: The correct answer is \( \boxed{\text{(a)}} \).