If `a_(1), a_(2), …, a_(n)` be an A.P. of positive terms, then `(a_(1) + a_(2n))/(sqrt(a_(1)) + sqrt(a_(2))) + (a_(2) + a_(2n - 1))/(sqrt(a_(2)) + sqrt(a_(3))) + ...+ (a_(n) + a_(n + 1))/(sqrt(a_(n)) + sqrt(a_(n + 1)))` is equal to

Options

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

Answer

To solve the given problem, we need to evaluate the expression:

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

where \( a_1, a_2, \ldots, a_n \) are terms of an arithmetic progression (A.P.) with positive terms.

### Step 1: Identify the terms of the A.P.
Let the first term of the A.P. be \( a \) and the common difference be \( d \). Then we can express the terms as:
- \( a_1 = a \)
- \( a_2 = a + d \)
- \( a_3 = a + 2d \)
- ...
- \( a_n = a + (n-1)d \)
- \( a_{2n} = a + (2n-1)d \)
- \( a_{n+1} = a + nd \)

### Step 2: Substitute the terms into the expression
We can rewrite the terms in the expression \( S \) using the definitions from Step 1:

\[ S = \frac{a + (2n-1)d}{\sqrt{a} + \sqrt{a + d}} + \frac{(a + d) + (2n-2)d}{\sqrt{a + d} + \sqrt{a + 2d}} + \ldots + \frac{(a + (n-1)d) + (a + nd)}{\sqrt{a + (n-1)d} + \sqrt{a + nd}} \]

### Step 3: Simplify each term
Notice that each term can be simplified using the identity:

\[ \frac{x + y}{\sqrt{x} + \sqrt{y}} = \frac{(\sqrt{x} + \sqrt{y})^2}{\sqrt{x} + \sqrt{y}} = \sqrt{x} + \sqrt{y} \]

Thus, we can rewrite each term:

\[ \frac{a_k + a_{k+1}}{\sqrt{a_k} + \sqrt{a_{k+1}}} = \sqrt{a_k} + \sqrt{a_{k+1}} \]

### Step 4: Sum the simplified terms
Now, we can sum the simplified terms:

\[ S = \sum_{k=1}^{n} (\sqrt{a_k} + \sqrt{a_{k+1}}) \]

This can be rearranged as:

\[ S = \sqrt{a_1} + \sqrt{a_2} + \sqrt{a_2} + \sqrt{a_3} + \ldots + \sqrt{a_n} + \sqrt{a_{n+1}} \]

### Step 5: Combine like terms
Notice that \( \sqrt{a_2}, \sqrt{a_3}, \ldots, \sqrt{a_n} \) appear twice. Thus, we can express \( S \) as:

\[ S = \sqrt{a_1} + 2(\sqrt{a_2} + \sqrt{a_3} + \ldots + \sqrt{a_n}) + \sqrt{a_{n+1}} \]

### Step 6: Conclusion
The final result can be expressed in terms of the first and last terms of the A.P. and the sum of the square roots of the intermediate terms.

Thus, the value of the expression is:

\[ S = \frac{a_1 + a_{2n}}{d} \]