Let `S` be the sum, `P` the product, and `R` the sum of reciprocals of `n` terms in a G.P. Prove that `P^2R^n=S^ndot`

Explanation

To prove that \( P^2 R^n = S^n \) for a geometric progression (G.P.), we start by defining the terms involved:

1. **Let \( a \)** be the first term of the G.P.
2. **Let \( r \)** be the common ratio.
3. **Let \( n \)** be the number of terms.

### Step 1: Find the sum \( S \) of the first \( n \) terms of the G.P.

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