If a, b, c be respectively the pth, qth and rth terms of an H.P., then `bc(q - r) + ca(r - p) + ab( p - q) =`

Options

A. 0
B. 1
C. -1
D. none

Answer

To solve the problem, we need to show that \( bc(q - r) + ca(r - p) + ab(p - q) = 0 \) given that \( a, b, c \) are the \( p \)-th, \( q \)-th, and \( r \)-th terms of a Harmonic Progression (H.P.).

### Step-by-Step Solution:

1. **Understanding Harmonic Progression**:
- If \( a, b, c \) are in H.P., then their reciprocals \( \frac{1}{a}, \frac{1}{b}, \frac{1}{c} \) are in Arithmetic Progression (A.P.).
- This means that:
\[ 2\frac{1}{b} = \frac{1}{a} + \frac{1}{c} \]

2. **Expressing Terms**:
- The \( p \)-th term of H.P. can be expressed as:
\[ a = \frac{1}{A + (p-1)D} \]
- The \( q \)-th term:
\[ b = \frac{1}{A + (q-1)D} \]
- The \( r \)-th term:
\[ c = \frac{1}{A + (r-1)D} \]
- Here, \( A \) and \( D \) are constants related to the H.P.

3. **Setting Up the Equation**:
- We need to evaluate \( bc(q - r) + ca(r - p) + ab(p - q) \).
- Substitute \( a, b, c \) into the equation:
\[ \frac{1}{A + (q-1)D} \cdot \frac{1}{A + (r-1)D} (q - r) + \frac{1}{A + (r-1)D} \cdot \frac{1}{A + (p-1)D} (r - p) + \frac{1}{A + (p-1)D} \cdot \frac{1}{A + (q-1)D} (p - q) \]

4. **Combining the Terms**:
- Combine the fractions:
\[ \frac{(q - r)}{(A + (q-1)D)(A + (r-1)D)} + \frac{(r - p)}{(A + (r-1)D)(A + (p-1)D)} + \frac{(p - q)}{(A + (p-1)D)(A + (q-1)D)} \]

5. **Finding a Common Denominator**:
- The common denominator will be \( (A + (p-1)D)(A + (q-1)D)(A + (r-1)D) \).
- Rewrite each term with the common denominator.

6. **Simplifying the Expression**:
- After simplification, you will find that the numerator becomes zero, leading to:
\[ bc(q - r) + ca(r - p) + ab(p - q) = 0 \]

### Final Result:
Thus, we conclude that:
\[ bc(q - r) + ca(r - p) + ab(p - q) = 0 \]