Suppose x and y are real numbers such that `-1 lt x lt y lt 1` Let G be the sum of the geometric series whose first term is x and whose common ratio is y, and let G' be the sum of the geomateric series whose first term is y and common ratio is x. If G = G' then the value of (x + y) is

Options

A. 1
B. `-1`
C. `2`
D. `-2`

Answer

To solve the problem, we need to find the value of \( x + y \) given the conditions and the equality of two geometric series sums \( G \) and \( G' \).

### Step-by-Step Solution:

1. **Define the sums of the geometric series**:
- The sum \( G \) of the geometric series with first term \( x \) and common ratio \( y \) is given by:
\[ G = \frac{x}{1 - y} \quad \text{(for } |y| < 1\text{)} \]
- The sum \( G' \) of the geometric series with first term \( y \) and common ratio \( x \) is given by:
\[ G' = \frac{y}{1 - x} \quad \text{(for } |x| < 1\text{)} \]

2. **Set the sums equal to each other**:
\[ G = G' \implies \frac{x}{1 - y} = \frac{y}{1 - x} \]

3. **Cross-multiply to eliminate the fractions**:
\[ x(1 - x) = y(1 - y) \]
Expanding both sides gives:
\[ x - x^2 = y - y^2 \]

4. **Rearrange the equation**:
\[ x - y = x^2 - y^2 \]
This can be factored using the difference of squares:
\[ x - y = (x - y)(x + y) \]

5. **Factor out \( x - y \)**:
If \( x \neq y \), we can divide both sides by \( x - y \):
\[ 1 = x + y \]
If \( x = y \), it contradicts the condition \( x < y \). Thus, we conclude:
\[ x + y = 1 \]

6. **Final answer**:
Therefore, the value of \( x + y \) is:
\[ \boxed{1} \]