If the pth, qth and rth terms of a G.P. are a,b and c, respectively. Prove that
`a^(q-r)b^(r-p)c^(p-q)=1`.

Explanation

To prove that \( a^{(q-r)} b^{(r-p)} c^{(p-q)} = 1 \), where \( a, b, c \) are the \( p \)-th, \( q \)-th, and \( r \)-th terms of a geometric progression (G.P.), we can follow these steps:

### Step 1: Define the terms of the G.P.
Let the first term of the G.P. be \( A \) and the common ratio be \( R \). Then, the terms can be expressed as:
- \( a = A \cdot R^{p-1} \) (the \( p \)-th term)
- \( b = A \cdot R^{q-1} \) (the \( q \)-th term)
- \( c = A \cdot R^{r-1} \) (the \( r \)-th term)

### Step 2: Substitute the terms into the expression
We need to substitute \( a, b, c \) into the expression \( a^{(q-r)} b^{(r-p)} c^{(p-q)} \):
\[ a^{(q-r)} = (A \cdot R^{p-1})^{(q-r)} = A^{(q-r)} \cdot R^{(p-1)(q-r)} \]
\[ b^{(r-p)} = (A \cdot R^{q-1})^{(r-p)} = A^{(r-p)} \cdot R^{(q-1)(r-p)} \]
\[ c^{(p-q)} = (A \cdot R^{r-1})^{(p-q)} = A^{(p-q)} \cdot R^{(r-1)(p-q)} \]

### Step 3: Combine the expressions
Now, combine these three expressions:
\[ a^{(q-r)} b^{(r-p)} c^{(p-q)} = A^{(q-r)} \cdot R^{(p-1)(q-r)} \cdot A^{(r-p)} \cdot R^{(q-1)(r-p)} \cdot A^{(p-q)} \cdot R^{(r-1)(p-q)} \]

### Step 4: Simplify the powers of \( A \)
Combining the powers of \( A \):
\[ A^{(q-r) + (r-p) + (p-q)} = A^{0} = 1 \]

### Step 5: Simplify the powers of \( R \)
Now, combine the powers of \( R \):
\[ R^{(p-1)(q-r) + (q-1)(r-p) + (r-1)(p-q)} \]

### Step 6: Expand and simplify the exponent of \( R \)
Expanding the exponent:
\[ = (p-1)(q-r) + (q-1)(r-p) + (r-1)(p-q) \]
\[ = pq - pr - q + r + qr - qp - r + p + rp - rq - p + q \]
Notice that terms will cancel out:
- \( pq \) and \( -qp \) cancel
- \( -pr \) and \( rp \) cancel
- \( -q \) and \( q \) cancel
- \( -r \) and \( r \) cancel
- \( -p \) and \( p \) cancel

Thus, the entire exponent simplifies to \( 0 \):
\[ R^{0} = 1 \]

### Step 7: Conclusion
Putting it all together:
\[ a^{(q-r)} b^{(r-p)} c^{(p-q)} = 1 \cdot 1 = 1 \]

Therefore, we have proved that:
\[ a^{(q-r)} b^{(r-p)} c^{(p-q)} = 1 \]