If the `p^(th), q^(th) and r^(th)` terms of an A.P. be a, b, c respectively, then
`a (q - r) + b(r - p) + c(p - q) =`

Options

A. 0
B. 2
C. p + q + r
D. pqr

Answer

To solve the problem, we need to express the terms of the arithmetic progression (A.P.) and then manipulate the given expression.

### Step-by-step Solution:

1. **Define the A.P. Terms**:
Let the first term of the A.P. be \( A \) and the common difference be \( D \).
The \( p^{th} \), \( q^{th} \), and \( r^{th} \) terms of the A.P. can be expressed as:
\[ a = A + (p-1)D \]
\[ b = A + (q-1)D \]
\[ c = A + (r-1)D \]

2. **Rearranging the Terms**:
We can rearrange these equations to isolate \( A \):
\[ A = a - (p-1)D \]
\[ A = b - (q-1)D \]
\[ A = c - (r-1)D \]

3. **Substituting into the Expression**:
We need to evaluate the expression:
\[ a(q - r) + b(r - p) + c(p - q) \]
Substituting the values of \( a \), \( b \), and \( c \):
\[ = (A + (p-1)D)(q - r) + (A + (q-1)D)(r - p) + (A + (r-1)D)(p - q) \]

4. **Expanding the Expression**:
Expanding each term:
\[ = A(q - r) + (p-1)D(q - r) + A(r - p) + (q-1)D(r - p) + A(p - q) + (r-1)D(p - q) \]

5. **Combining Like Terms**:
Combine the terms involving \( A \):
\[ = A[(q - r) + (r - p) + (p - q)] + D[(p-1)(q - r) + (q-1)(r - p) + (r-1)(p - q)] \]
Notice that the first part simplifies to:
\[ (q - r) + (r - p) + (p - q) = 0 \]
Thus, the entire expression simplifies to:
\[ = 0 + D[(p-1)(q - r) + (q-1)(r - p) + (r-1)(p - q)] \]

6. **Evaluating the Remaining Expression**:
We need to evaluate:
\[ (p-1)(q - r) + (q-1)(r - p) + (r-1)(p - q) \]
This expression is also equal to 0. Therefore, the entire expression evaluates to:
\[ = 0 \]

### Final Answer:
Thus, we conclude that:
\[ a(q - r) + b(r - p) + c(p - q) = 0 \]