Show that the products of the corresponding terms of the sequence `a,ar, ar^(2),… ar^(n-1)` and `A,AR, AR^(2),…AR^(n-1)` from a G.P. and find the common ratio.

Explanation

To show that the products of the corresponding terms of the sequences \( a, ar, ar^2, \ldots, ar^{n-1} \) and \( A, AR, AR^2, \ldots, AR^{n-1} \) form a geometric progression (G.P.) and to find the common ratio, we can follow these steps:

### Step 1: Write down the sequences
The first sequence is:
- \( a, ar, ar^2, \ldots, ar^{n-1} \)

The second sequence is:
- \( A, AR, AR^2, \ldots, AR^{n-1} \)

### Step 2: Find the products of corresponding terms
Now, we will find the product of the corresponding terms from both sequences:
- The first term: \( a \cdot A \)
- The second term: \( ar \cdot AR = aA \cdot rR \)
- The third term: \( ar^2 \cdot AR^2 = aA \cdot r^2R^2 \)
- Continuing this way, the \( k^{th} \) term will be: \( ar^{k-1} \cdot AR^{k-1} = aA \cdot r^{k-1}R^{k-1} \)

Thus, the products of corresponding terms form the sequence:
- \( aA, aA \cdot rR, aA \cdot r^2R^2, \ldots, aA \cdot r^{n-1}R^{n-1} \)

### Step 3: Identify the common ratio
To show that this sequence is a G.P., we need to find the common ratio between consecutive terms.

The first term is \( aA \) and the second term is \( aA \cdot rR \). The common ratio \( R \) can be calculated as follows:
\[ R = \frac{\text{Second term}}{\text{First term}} = \frac{aA \cdot rR}{aA} = rR \]

Now, let's check the ratio between the second term and the first term:
\[ \text{Second term} = aA \cdot rR \]
\[ \text{Third term} = aA \cdot r^2R^2 \]
The common ratio between the second and third terms is:
\[ \frac{\text{Third term}}{\text{Second term}} = \frac{aA \cdot r^2R^2}{aA \cdot rR} = \frac{r^2R^2}{rR} = rR \]

Thus, the common ratio between any two consecutive terms is \( rR \).

### Conclusion
The products of the corresponding terms of the sequences \( a, ar, ar^2, \ldots, ar^{n-1} \) and \( A, AR, AR^2, \ldots, AR^{n-1} \) form a G.P. with a common ratio of \( rR \).