If `a_1 , a_2 , a_3, …."" a_n` are in A.P. where `a_i gt 0` for all `i` , then the value of
`(1)/(sqrt(a_1) + sqrt(a_2)) + (1)/(sqrt(a_2) + sqrt(a_3)) + … + (1)/(sqrt(a_(n-1)) + sqrt(a_n))` :

Options

A. a. `(n-1)/(sqrt(a_1) + sqrt(a_2))`
B. b. `(n+1)/(sqrt(a_1) + sqrt(a_2))`
C. c. `(n)/(sqrt(a_1) - sqrt(a_2))`
D. d. none of these

Answer

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

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

where \( a_1, a_2, a_3, \ldots, a_n \) are in Arithmetic Progression (A.P.) and \( a_i > 0 \) for all \( i \).

### Step 1: Understand the properties of A.P.
Since \( a_1, a_2, a_3, \ldots, a_n \) are in A.P., we can express them as:
\[ a_k = a_1 + (k-1)d \quad \text{for } k = 1, 2, \ldots, n \]
where \( d \) is the common difference.

### Step 2: Rewrite the terms in the sum
We can express the terms in the sum using the definition of \( a_k \):
\[ S = \sum_{i=1}^{n-1} \frac{1}{\sqrt{a_i} + \sqrt{a_{i+1}}} \]

### Step 3: Rationalize each term
To simplify each term, we can rationalize:
\[ \frac{1}{\sqrt{a_i} + \sqrt{a_{i+1}}} \cdot \frac{\sqrt{a_{i+1}} - \sqrt{a_i}}{\sqrt{a_{i+1}} - \sqrt{a_i}} = \frac{\sqrt{a_{i+1}} - \sqrt{a_i}}{a_{i+1} - a_i} \]

Since \( a_{i+1} - a_i = d \), we have:
\[ \frac{\sqrt{a_{i+1}} - \sqrt{a_i}}{d} \]

### Step 4: Substitute back into the sum
Now substituting this back into the sum:
\[ S = \sum_{i=1}^{n-1} \frac{\sqrt{a_{i+1}} - \sqrt{a_i}}{d} \]

### Step 5: Recognize the telescoping nature
This sum is telescoping:
\[ S = \frac{1}{d} \left( \sqrt{a_n} - \sqrt{a_1} \right) \]

### Step 6: Final expression
Thus, we can express \( S \) as:
\[ S = \frac{\sqrt{a_n} - \sqrt{a_1}}{d} \]

### Step 7: Substitute \( d \) back in terms of \( a_1 \) and \( a_n \)
Since \( d = a_2 - a_1 \) and we can express \( a_n \) in terms of \( a_1 \) and \( d \):
\[ a_n = a_1 + (n-1)d \]

### Final Result
The final expression for \( S \) is:
\[ S = \frac{\sqrt{a_n} - \sqrt{a_1}}{d} \]