If S be the sum, P the product and R the sum of the reciprocals of n terms of a G.P., then `((S)/(R))^(n) =`

Options

A. P
B. `P^(2)`
C. `P^(3)`
D. `sqrt(P)`

Answer

To solve the problem, we need to find the expression for \(\left(\frac{S}{R}\right)^n\) where \(S\) is the sum, \(P\) is the product, and \(R\) is the sum of the reciprocals of \(n\) terms of a geometric progression (G.P.).

### Step 1: Calculate the Sum \(S\)
The sum \(S\) of the first \(n\) terms of a G.P. with the first term \(A\) and common ratio \(R\) is given by the formula:

\[ S = A + AR + AR^2 + \ldots + AR^{n-1} = A \frac{1 - R^n}{1 - R} \quad \text{(for } R \neq 1\text{)} \]

### Step 2: Calculate the Product \(P\)
The product \(P\) of the first \(n\) terms of the G.P. is:

\[ P = A \cdot AR \cdot AR^2 \cdots AR^{n-1} = A^n \cdot R^{0 + 1 + 2 + \ldots + (n-1)} = A^n \cdot R^{\frac{(n-1)n}{2}} \]

### Step 3: Calculate the Sum of Reciprocals \(R\)
The sum of the reciprocals of the first \(n\) terms of the G.P. is:

\[ R = \frac{1}{A} + \frac{1}{AR} + \frac{1}{AR^2} + \ldots + \frac{1}{AR^{n-1}} = \frac{1}{A} \left(1 + \frac{1}{R} + \frac{1}{R^2} + \ldots + \frac{1}{R^{n-1}}\right) \]

This is a geometric series with first term \(1\) and common ratio \(\frac{1}{R}\):

\[ R = \frac{1}{A} \cdot \frac{1 - \left(\frac{1}{R}\right)^n}{1 - \frac{1}{R}} = \frac{1}{A} \cdot \frac{R^n - 1}{R^n(R - 1)} \]

### Step 4: Calculate \(\frac{S}{R}\)
Now we can find \(\frac{S}{R}\):

\[ \frac{S}{R} = \frac{A \frac{1 - R^n}{1 - R}}{\frac{1}{A} \cdot \frac{R^n - 1}{R^n(R - 1)}} \]

This simplifies to:

\[ \frac{S}{R} = A^2 \cdot \frac{(1 - R^n)(R^n(R - 1))}{(R^n - 1)(1 - R)} \]

### Step 5: Raise to the Power of \(n\)
Finally, we need to raise \(\frac{S}{R}\) to the power of \(n\):

\[ \left(\frac{S}{R}\right)^n = \left(A^2 \cdot \frac{(1 - R^n)(R^n(R - 1))}{(R^n - 1)(1 - R)}\right)^n \]

### Final Result
Thus, we have:

\[ \left(\frac{S}{R}\right)^n = A^{2n} \cdot \left(\frac{(1 - R^n)(R^n(R - 1))}{(R^n - 1)(1 - R)}\right)^n \]