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_(nrarroo)sqrtd/n(1/(sqrta_1+sqrta_2)+1/(sqrta_2+sqrta_3)+......+1/(sqrta_(n-1)+sqrta_n))`

Options

A. `1/(sqrtd)`
B. `sqrtd`
C. `1`
D. `0`

Answer

To solve the limit problem given, we start with the expression:

\[ \lim_{n \to \infty} \frac{\sqrt{d}}{n} \left( \frac{1}{\sqrt{a_1} + \sqrt{a_2}} + \frac{1}{\sqrt{a_2} + \sqrt{a_3}} + \ldots + \frac{1}{\sqrt{a_{n-1}} + \sqrt{a_n}} \right) \]

### Step 1: Define the terms of the arithmetic progression
Let the first term \( a_1 = a \) (where \( a > 0 \)) and the common difference \( d \). The \( n \)-th term of the arithmetic progression can be expressed as:
\[ a_n = a + (n-1)d \]

### Step 2: Write the sum in the limit
The sum inside the limit can be rewritten as:
\[ \sum_{k=1}^{n-1} \frac{1}{\sqrt{a_k} + \sqrt{a_{k+1}}} \]
where \( a_k = a + (k-1)d \) and \( a_{k+1} = a + kd \).

### Step 3: Simplify each term
Now, we simplify each term:
\[ \sqrt{a_k} + \sqrt{a_{k+1}} = \sqrt{a + (k-1)d} + \sqrt{a + kd} \]
Using the approximation for large \( n \), we can factor out \( \sqrt{d} \):
\[ \sqrt{a_k} + \sqrt{a_{k+1}} \approx \sqrt{d} \left( \sqrt{\frac{a + (k-1)d}{d}} + \sqrt{\frac{a + kd}{d}} \right) \]

### Step 4: Analyze the limit as \( n \to \infty \)
As \( n \) approaches infinity, the terms \( \sqrt{a + (k-1)d} \) and \( \sqrt{a + kd} \) both grow large. Thus, we can approximate:
\[ \frac{1}{\sqrt{a_k} + \sqrt{a_{k+1}}} \approx \frac{1}{2\sqrt{d} \sqrt{k d}} = \frac{1}{2\sqrt{d^2 k}} = \frac{1}{2d\sqrt{k}} \]

### Step 5: Substitute back into the sum
Now, substituting this back into our sum gives:
\[ \sum_{k=1}^{n-1} \frac{1}{\sqrt{a_k} + \sqrt{a_{k+1}}} \approx \sum_{k=1}^{n-1} \frac{1}{2d\sqrt{k}} \]

### Step 6: Evaluate the sum
The sum \( \sum_{k=1}^{n-1} \frac{1}{\sqrt{k}} \) behaves like \( 2\sqrt{n} \) for large \( n \). Thus:
\[ \sum_{k=1}^{n-1} \frac{1}{\sqrt{k}} \approx 2\sqrt{n} \]

### Step 7: Final limit calculation
Substituting this back into our limit expression, we have:
\[ \lim_{n \to \infty} \frac{\sqrt{d}}{n} \cdot \frac{1}{2d} \cdot 2\sqrt{n} = \lim_{n \to \infty} \frac{\sqrt{d}}{n} \cdot \frac{2\sqrt{n}}{2d} = \lim_{n \to \infty} \frac{\sqrt{d}}{d} \cdot \frac{1}{\sqrt{n}} = 0 \]

Thus, the final answer is:
\[ \boxed{0} \]