If `a_r>0, r in N` and `a_1.a_2,....a_(2n)` are in A.P then `(a_1+a_2)/(sqrta_1+sqrta_2)+(a_2+a_(2n-1))/(sqrta_2+sqrta_3)+.....+(a_n+a_(n+1))/(sqrt a_n+sqrta_(n+1))=`

Options

A. n-1
B. `(n(a_(1)+a_(2n)))/(sqrt(a_(1))+sqrt(a_(n+1)))`
C. `(n-1)/(sqrt(a_(1))+sqrt(a_(n+1)))`
D. none of these

Explanation

To solve the problem, we need to find the value of the expression:

\[ \frac{a_1 + a_2}{\sqrt{a_1} + \sqrt{a_2}} + \frac{a_2 + a_{2n-1}}{\sqrt{a_2} + \sqrt{a_3}} + \ldots + \frac{a_n + a_{n+1}}{\sqrt{a_n} + \sqrt{a_{n+1}}} \]

Given that \( a_1, a_2, \ldots, a_{2n} \) are in Arithmetic Progression (A.P.), we can denote the first term as \( a_1 \) and the common difference as \( d \). Thus, we have:

\[ a_r = a_1 + (r-1)d \quad \text{for } r = 1, 2, \ldots, 2n \]

### Step 1: Identify the Terms in the Expression

The terms in the expression can be rewritten using the general formula for the terms in A.P.:

- \( a_1 = a_1 \)
- \( a_2 = a_1 + d \)
- \( a_3 = a_1 + 2d \)
- ...
- \( a_n = a_1 + (n-1)d \)
- \( a_{n+1} = a_1 + nd \)
- ...
- \( a_{2n} = a_1 + (2n-1)d \)

### Step 2: Rewrite Each Fraction

Now, we can rewrite each term in the expression:

\[ \frac{a_1 + a_2}{\sqrt{a_1} + \sqrt{a_2}} = \frac{a_1 + (a_1 + d)}{\sqrt{a_1} + \sqrt{a_1 + d}} = \frac{2a_1 + d}{\sqrt{a_1} + \sqrt{a_1 + d}} \]

Continuing this process for each term, we will have:

\[ \frac{a_2 + a_{2n-1}}{\sqrt{a_2} + \sqrt{a_{2n-1}}} = \frac{(a_1 + d) + (a_1 + (2n-2)d)}{\sqrt{a_1 + d} + \sqrt{a_1 + (2n-2)d}} \]

And so forth for each term.

### Step 3: Simplify the Expression

We can see that each term can be simplified using the identity:

\[ \frac{x+y}{\sqrt{x}+\sqrt{y}} = \sqrt{x} + \sqrt{y} \]

Thus, we can rewrite the entire expression as:

\[ \sum_{r=1}^{n} \left( \sqrt{a_r} + \sqrt{a_{r+1}} \right) \]

### Step 4: Calculate the Final Sum

The sum can be calculated as:

\[ \sum_{r=1}^{n} \left( \sqrt{a_r} + \sqrt{a_{r+1}} \right) = \sqrt{a_1} + \sqrt{a_2} + \sqrt{a_2} + \sqrt{a_3} + \ldots + \sqrt{a_n} + \sqrt{a_{n+1}} \]

This results in:

\[ = \sqrt{a_1} + 2\sum_{r=2}^{n} \sqrt{a_r} + \sqrt{a_{n+1}} \]

### Final Result

Thus, the final result is:

\[ \frac{n(a_1 + a_{2n})}{\sqrt{a_1} + \sqrt{a_{2n}}} \]