If `a_1, a_2, a_3,.....a_n` are in H.P. and `a_1 a_2+a_2 a_3+a_3 a_4+.......a_(n-1) a_n=ka_1 a_n`, then k is equal to

Options

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

Explanation

To solve the problem, we need to find the value of \( k \) given that \( a_1, a_2, a_3, \ldots, a_n \) are in Harmonic Progression (H.P.) and that the equation \( a_1 a_2 + a_2 a_3 + a_3 a_4 + \ldots + a_{n-1} a_n = k a_1 a_n \) holds.

### Step-by-step Solution:

1. **Understanding H.P. and A.P. Relation**:
Since \( a_1, a_2, a_3, \ldots, a_n \) are in H.P., the reciprocals \( \frac{1}{a_1}, \frac{1}{a_2}, \frac{1}{a_3}, \ldots, \frac{1}{a_n} \) are in Arithmetic Progression (A.P.).

2. **Common Difference of A.P.**:
Let the common difference of the A.P. be \( d \). Thus, we can express the differences:
\[ \frac{1}{a_2} - \frac{1}{a_1} = d \quad \text{and} \quad \frac{1}{a_3} - \frac{1}{a_2} = d \]
This gives us:
\[ \frac{a_1 - a_2}{a_1 a_2} = d \quad \text{(1)} \]
\[ \frac{a_2 - a_3}{a_2 a_3} = d \quad \text{(2)} \]

3. **Finding the nth Term of A.P.**:
The nth term of the A.P. can be expressed as:
\[ \frac{1}{a_n} = \frac{1}{a_1} + (n-1)d \]
Rearranging gives:
\[ a_n = \frac{a_1}{1 + (n-1)a_1 d} \quad \text{(3)} \]

4. **Summing the Products**:
The left-hand side of the equation given in the problem can be expressed as:
\[ a_1 a_2 + a_2 a_3 + \ldots + a_{n-1} a_n \]
Using equations (1) and (2), we can express each product in terms of \( d \) and the \( a_i \).

5. **Expressing the Left-Hand Side**:
We can write:
\[ a_1 a_2 = \frac{a_1 (a_1 - a_2)}{d} + \frac{a_2 (a_2 - a_3)}{d} + \ldots + \frac{a_{n-1} (a_{n-1} - a_n)}{d} \]
Factoring out \( \frac{1}{d} \) gives:
\[ \frac{1}{d} \left( a_1 - a_n \right) = k a_1 a_n \]

6. **Substituting Known Values**:
From (3), we substitute \( a_n \) back into the equation:
\[ \frac{(n-1) a_1 a_n}{d} = k a_1 a_n \]

7. **Solving for \( k \)**:
Dividing both sides by \( a_1 a_n \) (assuming \( a_1, a_n \neq 0 \)):
\[ \frac{n-1}{d} = k \]
Since \( d \) is the common difference, we can express \( k \) as:
\[ k = n - 1 \]

### Final Result:
Thus, the value of \( k \) is:
\[ \boxed{n - 1} \]