Let `a_n, n in N` is an A.P with common difference `d` and all whose terms are non-zero. If n approaches infinity, then the sum `1/(a_1a_2)+1/(a_2a_3)+....1/(a_na_(n+1))` will approach

Options

A. `(1)/(a_(1)d)`
B. `(2)/(a_(1)d)`
C. `(1)/(2a_(1)d)`
D. `a_(1)d`

Answer

To solve the problem, we need to evaluate the limit of the sum

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

as \( n \) approaches infinity, where \( a_n \) is the \( n \)-th term of an arithmetic progression (A.P.) with first term \( a_1 \) and common difference \( d \).

### Step-by-step Solution:

1. **Define the terms of the A.P.**:
The \( n \)-th term of the A.P. can be expressed as:
\[ a_n = a_1 + (n-1)d \]
Therefore, the terms are:
- \( a_1 = a_1 \)
- \( a_2 = a_1 + d \)
- \( a_3 = a_1 + 2d \)
- \( \ldots \)
- \( a_{n+1} = a_1 + nd \)

2. **Rewrite the sum \( S_n \)**:
We can express \( S_n \) as:
\[ S_n = \sum_{k=1}^{n} \frac{1}{a_k a_{k+1}} = \frac{1}{a_1 a_2} + \frac{1}{a_2 a_3} + \ldots + \frac{1}{a_n a_{n+1}} \]

3. **Use the identity for the sum**:
We can rewrite each term:
\[ \frac{1}{a_k a_{k+1}} = \frac{1}{d} \left( \frac{1}{a_k} - \frac{1}{a_{k+1}} \right) \]
This allows us to express \( S_n \) as:
\[ S_n = \frac{1}{d} \left( \frac{1}{a_1} - \frac{1}{a_{n+1}} \right) \]

4. **Substitute \( a_{n+1} \)**:
Substitute \( a_{n+1} = a_1 + nd \) into the expression:
\[ S_n = \frac{1}{d} \left( \frac{1}{a_1} - \frac{1}{a_1 + nd} \right) \]

5. **Evaluate the limit as \( n \to \infty \)**:
As \( n \) approaches infinity, \( a_{n+1} \) approaches infinity, thus:
\[ \frac{1}{a_{n+1}} \to 0 \]
Therefore, we have:
\[ S_n \to \frac{1}{d} \left( \frac{1}{a_1} - 0 \right) = \frac{1}{d a_1} \]

### Final Result:
Thus, the sum \( S_n \) approaches:
\[ \boxed{\frac{1}{d a_1}} \]