If the first and the nth term of a G.P. are a and b, respectively, and if P is the product of n terms , prove that `P^(2)=(ab)^(n)`.

Explanation

To prove that \( P^2 = (ab)^n \) where \( P \) is the product of the first \( n \) terms of a geometric progression (G.P.) with first term \( a \) and \( n \)-th term \( b \), we can follow these steps:

### Step 1: Understand the terms of the G.P.
In a G.P., the first term is \( a \) and the \( n \)-th term can be expressed as:
\[ b = a \cdot r^{n-1} \]
where \( r \) is the common ratio of the G.P.

### Step 2: Write the product of the first \( n \) terms
The first \( n \) terms of the G.P. are:
\[ a, ar, ar^2, \ldots, ar^{n-1} \]
The product \( P \) of these \( n \) terms is:
\[ P = a \cdot ar \cdot ar^2 \cdots ar^{n-1} \]

### Step 3: Factor out \( a \)
We can factor \( a \) out of the product:
\[ P = a^n \cdot (1 \cdot r \cdot r^2 \cdots r^{n-1}) \]

### Step 4: Simplify the product of the ratios
The product of the ratios can be simplified using the formula for the product of a geometric series:
\[ 1 \cdot r \cdot r^2 \cdots r^{n-1} = r^{0 + 1 + 2 + \ldots + (n-1)} = r^{\frac{(n-1)n}{2}} \]
Thus, we have:
\[ P = a^n \cdot r^{\frac{(n-1)n}{2}} \]

### Step 5: Substitute \( r \) in terms of \( a \) and \( b \)
From the expression for \( b \), we can express \( r \):
\[ r^{n-1} = \frac{b}{a} \implies r = \left(\frac{b}{a}\right)^{\frac{1}{n-1}} \]

### Step 6: Calculate \( P^2 \)
Now, substituting \( r \) back into the expression for \( P \):
\[ P = a^n \cdot \left(\left(\frac{b}{a}\right)^{\frac{1}{n-1}}\right)^{\frac{(n-1)n}{2}} = a^n \cdot \left(\frac{b}{a}\right)^{\frac{n}{2}} \]
This simplifies to:
\[ P = a^n \cdot \frac{b^{\frac{n}{2}}}{a^{\frac{n}{2}}} = a^{n - \frac{n}{2}} \cdot b^{\frac{n}{2}} = a^{\frac{n}{2}} \cdot b^{\frac{n}{2}} \]

### Step 7: Square \( P \)
Now, squaring \( P \):
\[ P^2 = \left(a^{\frac{n}{2}} \cdot b^{\frac{n}{2}}\right)^2 = a^n \cdot b^n = (ab)^n \]

### Conclusion
Thus, we have proved that:
\[ P^2 = (ab)^n \]