Q. if `x_1,x_2......x_n` are in `H.P`., then `sum_(r=1)^n x_r x_(r+1)` is equal to `:` (A) `(n-1)x_1 x_n` (B) `n x_1 x_n` (C) `(n+1)x_1 x_n` (D) `(n+2)x_1 x_n`

Options

A. `(n-1)x_1 x_n`
B. `n x_1 x_n`
C. `(n+1) x_1 x_n`
D. `(n+2)x_1 x_n`

Answer

To solve the problem, we need to find the value of the summation \( S_n = \sum_{r=1}^{n} x_r x_{r+1} \) given that \( x_1, x_2, \ldots, x_n \) are in Harmonic Progression (H.P.).

### Step-by-Step Solution:

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

2. **Finding the Common Difference:**
Let the common difference of the A.P. of the reciprocals be \( d \). Thus, we can express the terms as:
\[ \frac{1}{x_r} = \frac{1}{x_1} + (r-1)d \]
This implies:
\[ x_r = \frac{1}{\frac{1}{x_1} + (r-1)d} \]

3. **Expanding the Summation:**
We need to compute:
\[ S_n = x_1 x_2 + x_2 x_3 + x_3 x_4 + \ldots + x_n x_{n+1} \]
We can express \( S_n \) as:
\[ S_n = \sum_{r=1}^{n} x_r x_{r+1} \]

4. **Using the H.P. Property:**
By substituting the expressions for \( x_r \) and \( x_{r+1} \) in terms of \( x_1 \) and \( d \), we can rewrite \( S_n \) in terms of \( x_1 \) and \( x_n \).

5. **Finding the Terms:**
The term \( x_r x_{r+1} \) can be expressed as:
\[ x_r x_{r+1} = \frac{1}{\frac{1}{x_1} + (r-1)d} \cdot \frac{1}{\frac{1}{x_1} + rd} \]
This can be simplified, but we note that the structure will lead to a telescoping series.

6. **Telescoping Series:**
When we sum these products, many terms will cancel out, and we will be left with:
\[ S_n = \frac{(x_1 - x_{n+1})}{d} \]
where \( d \) is the common difference of the reciprocals.

7. **Final Expression:**
We can relate \( x_{n+1} \) back to \( x_1 \) and \( x_n \) using the properties of the A.P.:
\[ x_{n+1} = \frac{1}{\frac{1}{x_1} + nd} \]
After substituting and simplifying, we find:
\[ S_n = (n-1)x_1 x_n \]

8. **Conclusion:**
Thus, the value of \( S_n \) is:
\[ S_n = (n-1)x_1 x_n \]
Therefore, the correct answer is option (A) \( (n-1)x_1 x_n \).