Show that the ratio of the sum of first n terms of a G.P. to the sum of terms from (n+1)th to (2n)th term is `(1)/(r^(n))`.

Explanation

To show that the ratio of the sum of the first n terms of a geometric progression (G.P.) to the sum of the terms from the (n+1)th to the 2n-th term is \( \frac{1}{r^n} \), we will follow these steps:

### Step 1: Sum of the first n terms of a G.P.

The sum \( S_n \) of the first n terms of a G.P. with first term \( a \) and common ratio \( r \) is given by the formula:

\[ S_n = \frac{a(1 - r^n)}{1 - r} \quad (r \neq 1) \]

### Step 2: Sum of the terms from (n+1)th to (2n)th

To find the sum of the terms from the (n+1)th to the 2n-th term, we can denote this sum as \( S_{n+1 \text{ to } 2n} \).

The sum of the first 2n terms is:

\[ S_{2n} = \frac{a(1 - r^{2n})}{1 - r} \]

The sum of the first n terms is \( S_n \) as calculated above. Therefore, the sum from the (n+1)th to the 2n-th term can be found by subtracting \( S_n \) from \( S_{2n} \):

\[ S_{n+1 \text{ to } 2n} = S_{2n} - S_n \]

Substituting the values we have:

\[ S_{n+1 \text{ to } 2n} = \frac{a(1 - r^{2n})}{1 - r} - \frac{a(1 - r^n)}{1 - r} \]

### Step 3: Simplifying the expression

Now, we can simplify \( S_{n+1 \text{ to } 2n} \):

\[ S_{n+1 \text{ to } 2n} = \frac{a(1 - r^{2n}) - a(1 - r^n)}{1 - r} \]

\[ = \frac{a(-r^{2n} + r^n)}{1 - r} \]

\[ = \frac{a r^n (1 - r^n)}{1 - r} \]

### Step 4: Finding the ratio

Now we need to find the ratio \( \frac{S_n}{S_{n+1 \text{ to } 2n}} \):

\[ \frac{S_n}{S_{n+1 \text{ to } 2n}} = \frac{\frac{a(1 - r^n)}{1 - r}}{\frac{a r^n (1 - r^n)}{1 - r}} \]

The \( a \) and \( (1 - r) \) terms cancel out:

\[ = \frac{1 - r^n}{r^n(1 - r^n)} \]

\[ = \frac{1}{r^n} \]

### Conclusion

Thus, we have shown that:

\[ \frac{S_n}{S_{n+1 \text{ to } 2n}} = \frac{1}{r^n} \]