If the pth, qth and rth terms of a G.P. be respectively , a,b and c then the value of `a^(q-r).b^(r-p).c^(p-q)` is :

Options

A. A. 0
B. B. 1
C. C. 3
D. D. `(abc)/(pqr)`

Answer

To solve the problem, we need to find the value of \( a^{(q-r)} \cdot b^{(r-p)} \cdot c^{(p-q)} \) given that \( a \), \( b \), and \( c \) are the \( p \)-th, \( q \)-th, and \( r \)-th terms of a geometric progression (G.P.).

### Step-by-Step Solution:

1. **Identify the Terms of the G.P.**:
- The \( p \)-th term \( a \) can be expressed as:
\[ a = A \cdot r^{p-1} \]
- The \( q \)-th term \( b \) can be expressed as:
\[ b = A \cdot r^{q-1} \]
- The \( r \)-th term \( c \) can be expressed as:
\[ c = A \cdot r^{r-1} \]
where \( A \) is the first term of the G.P. and \( r \) is the common ratio.

2. **Substitute the Terms into the Expression**:
- We need to substitute \( a \), \( b \), and \( c \) into the expression \( a^{(q-r)} \cdot b^{(r-p)} \cdot c^{(p-q)} \):
\[ a^{(q-r)} = (A \cdot r^{p-1})^{(q-r)} = A^{(q-r)} \cdot r^{(p-1)(q-r)} \]
\[ b^{(r-p)} = (A \cdot r^{q-1})^{(r-p)} = A^{(r-p)} \cdot r^{(q-1)(r-p)} \]
\[ c^{(p-q)} = (A \cdot r^{r-1})^{(p-q)} = A^{(p-q)} \cdot r^{(r-1)(p-q)} \]

3. **Combine the Expressions**:
- Now, we combine all the terms:
\[ a^{(q-r)} \cdot b^{(r-p)} \cdot c^{(p-q)} = A^{(q-r)} \cdot r^{(p-1)(q-r)} \cdot A^{(r-p)} \cdot r^{(q-1)(r-p)} \cdot A^{(p-q)} \cdot r^{(r-1)(p-q)} \]
- This simplifies to:
\[ A^{(q-r) + (r-p) + (p-q)} \cdot r^{(p-1)(q-r) + (q-1)(r-p) + (r-1)(p-q)} \]

4. **Simplify the Exponents**:
- The exponent of \( A \):
\[ (q-r) + (r-p) + (p-q) = 0 \]
- The exponent of \( r \):
- Expanding:
\[ (p-1)(q-r) + (q-1)(r-p) + (r-1)(p-q) \]
- This can be simplified to:
\[ pq - pr - qr + qp + qr - qp - rp + rq + rp - rq = 0 \]

5. **Final Result**:
- Therefore, we have:
\[ a^{(q-r)} \cdot b^{(r-p)} \cdot c^{(p-q)} = A^0 \cdot r^0 = 1 \]

### Conclusion:
The value of \( a^{(q-r)} \cdot b^{(r-p)} \cdot c^{(p-q)} \) is **1**.