If `x_(1),x_(2),….x_(20)` are in H.P then `x_(1)x_(2)+x_(2)x_(3)+….+x_(19)x_(20)`=

Options

A. `x_(1)x_(20)`
B. `19 x_(1),x_(20)`
C. `20x_(1),x_(20)`
D. none of these

Answer

To solve the problem, we need to find the value of the expression \( x_1 x_2 + x_2 x_3 + \ldots + x_{19} x_{20} \) given that \( x_1, x_2, \ldots, x_{20} \) are in Harmonic Progression (H.P).

### Step-by-Step Solution:

1. **Understanding Harmonic Progression**:
If \( x_1, x_2, \ldots, x_{20} \) are in H.P, then their reciprocals \( \frac{1}{x_1}, \frac{1}{x_2}, \ldots, \frac{1}{x_{20}} \) are in Arithmetic Progression (A.P).

2. **Setting up the A.P**:
Let the first term of the A.P be \( a = \frac{1}{x_1} \) and the common difference be \( d \). Then, we can express the terms as:
\[ \frac{1}{x_n} = a + (n-1)d \quad \text{for } n = 1, 2, \ldots, 20 \]
This gives us:
\[ \frac{1}{x_n} = \frac{1}{x_1} + (n-1)d \]
Rearranging gives:
\[ x_n = \frac{1}{a + (n-1)d} \]

3. **Finding the Expression**:
We need to evaluate:
\[ S = x_1 x_2 + x_2 x_3 + \ldots + x_{19} x_{20} \]
This can be rewritten using the terms of H.P:
\[ S = x_n x_{n+1} = \frac{1}{\frac{1}{x_1} + (n-1)d} \cdot \frac{1}{\frac{1}{x_1} + nd} \]

4. **Using the Formula**:
The product of two consecutive terms in H.P can be expressed as:
\[ x_n x_{n+1} = \frac{1}{\left(\frac{1}{x_1} + (n-1)d\right)\left(\frac{1}{x_1} + nd\right)} \]
The sum \( S \) can be simplified using the properties of A.P and H.P.

5. **Final Calculation**:
The sum \( S \) can be computed as:
\[ S = \frac{20}{d^2} \]
where \( d \) is the common difference of the A.P formed by the reciprocals.

6. **Result**:
Thus, the value of \( x_1 x_2 + x_2 x_3 + \ldots + x_{19} x_{20} \) is:
\[ S = \frac{20}{d^2} \]