If `a_1,a_2,a_3,...,a_(n+1)` are in A.P. , then `1/(a_1a_2)+1/(a_2a_3)....+1/(a_na_(n+1))` is

Options

A. `(n-1)/(a_(1)a_(n+1))`
B. `(1)/(a_(1)a_(n+1))`
C. `(n+1)/(a_(1)a_(n+1))`
D. `(n)/(a_(1)a_(n+1))`

Explanation

To solve the problem, we need to find the sum of the series given that \( a_1, a_2, a_3, \ldots, a_{n+1} \) are in arithmetic progression (A.P.).

### Step-by-Step Solution:

1. **Understanding the A.P. Terms**:
Since \( a_1, a_2, a_3, \ldots, a_{n+1} \) are in A.P., we can express the terms as:
\[ a_1 = a, \quad a_2 = a + d, \quad a_3 = a + 2d, \quad \ldots, \quad a_{n+1} = a + nd \]
where \( d \) is the common difference.

2. **Writing the Given Expression**:
We need to evaluate:
\[ S = \frac{1}{a_1 a_2} + \frac{1}{a_2 a_3} + \ldots + \frac{1}{a_n a_{n+1}} \]

3. **Substituting the Terms**:
Substituting the values of \( a_i \):
\[ S = \frac{1}{a(a + d)} + \frac{1}{(a + d)(a + 2d)} + \ldots + \frac{1}{(a + (n-1)d)(a + nd)} \]

4. **Multiplying and Dividing by \( d \)**:
To simplify, we multiply and divide each term by \( d \):
\[ S = \frac{d}{a(a + d)} \cdot \frac{1}{d} + \frac{d}{(a + d)(a + 2d)} \cdot \frac{1}{d} + \ldots + \frac{d}{(a + (n-1)d)(a + nd)} \cdot \frac{1}{d} \]
This gives:
\[ S = \frac{d}{d} \left( \frac{1}{a(a + d)} + \frac{1}{(a + d)(a + 2d)} + \ldots + \frac{1}{(a + (n-1)d)(a + nd)} \right) \]

5. **Rearranging the Terms**:
The expression can be rewritten as:
\[ S = \frac{1}{d} \left( \frac{(a + d) - a}{a(a + d)} + \frac{(a + 2d) - (a + d)}{(a + d)(a + 2d)} + \ldots + \frac{(a + nd) - (a + (n-1)d)}{(a + (n-1)d)(a + nd)} \right) \]

6. **Telescoping Series**:
Notice that this forms a telescoping series:
\[ S = \frac{1}{d} \left( \frac{1}{a} - \frac{1}{a + nd} \right) \]

7. **Final Simplification**:
Thus, we can simplify \( S \):
\[ S = \frac{1}{d} \left( \frac{1}{a} - \frac{1}{a + nd} \right) = \frac{1}{d} \cdot \frac{(a + nd) - a}{a(a + nd)} = \frac{nd}{a(a + nd)} \]

8. **Conclusion**:
Finally, we have:
\[ S = \frac{n}{a(a + nd)} \]
where \( a = a_1 \) and \( a + nd = a_{n+1} \).

### Final Answer:
\[ S = \frac{n}{a_1 a_{n+1}} \]