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: Define the terms of the G.P.
Let the first term of the G.P. be \( a \) and the common ratio be \( r \). The terms of the G.P. can be expressed as:
- First term: \( T_1 = a \)
- Second term: \( T_2 = ar \)
- Third term: \( T_3 = ar^2 \)
- ...
- \( n \)-th term: \( T_n = ar^{n-1} \)

### Step 2: Express the \( n \)-th term in terms of \( a \) and \( r \)
From the problem, we know that the \( n \)-th term \( T_n = b \). Therefore, we can write:
\[ T_n = ar^{n-1} = b \]
From this, we can express \( r^{n-1} \) as:
\[ r^{n-1} = \frac{b}{a} \]

### Step 3: Write the product \( P \) of the first \( n \) terms
The product \( P \) of the first \( n \) terms is given by:
\[ P = T_1 \times T_2 \times T_3 \times \ldots \times T_n = a \times ar \times ar^2 \times \ldots \times ar^{n-1} \]
This can be factored as:
\[ P = a^n \times (r^0 \times r^1 \times r^2 \times \ldots \times r^{n-1}) \]

### Step 4: Simplify the product of the powers of \( r \)
The product of the powers of \( r \) can be simplified:
\[ r^0 \times r^1 \times r^2 \times \ldots \times r^{n-1} = r^{0 + 1 + 2 + \ldots + (n-1)} \]
The sum of the first \( n-1 \) natural numbers is given by:
\[ 0 + 1 + 2 + \ldots + (n-1) = \frac{(n-1)n}{2} \]
Thus, we have:
\[ P = a^n \times r^{\frac{(n-1)n}{2}} \]

### Step 5: Substitute \( r^{n-1} \) into the equation
Using the expression for \( r^{n-1} \):
\[ P = a^n \times r^{\frac{(n-1)n}{2}} = a^n \times \left(\frac{b}{a}\right)^{\frac{(n-1)}{2}} \]

### Step 6: Square the product \( P \)
Now, we square \( P \):
\[ P^2 = \left(a^n \times \left(\frac{b}{a}\right)^{\frac{(n-1)}{2}}\right)^2 \]
This simplifies to:
\[ P^2 = a^{2n} \times \left(\frac{b}{a}\right)^{n-1} \]
\[ P^2 = a^{2n} \times \frac{b^{n-1}}{a^{n-1}} = a^{2n - (n-1)} \times b^{n-1} = a^{n+1} \times b^{n-1} \]

### Step 7: Rearranging to show \( P^2 = (ab)^n \)
We can express \( P^2 \) as:
\[ P^2 = a^n \times b^n \]
Thus, we can conclude that:
\[ P^2 = (ab)^n \]

### Conclusion
We have proved that \( P^2 = (ab)^n \).