For what values of x, the numbers `-(2)/(7),x,-(7)/(2)" are in G.P."?`

Explanation

To determine the values of \( x \) for which the numbers \( -\frac{2}{7}, x, -\frac{7}{2} \) are in geometric progression (G.P.), we can use the property of G.P. that states the ratio of consecutive terms is constant.

### Step-by-Step Solution:

1. **Identify the terms:**
Let the first term \( a_1 = -\frac{2}{7} \), the second term \( a_2 = x \), and the third term \( a_3 = -\frac{7}{2} \).

2. **Set up the G.P. condition:**
For three numbers to be in G.P., the ratio of the second term to the first term must equal the ratio of the third term to the second term. This can be expressed as:
\[ \frac{a_2}{a_1} = \frac{a_3}{a_2} \]
Substituting the values, we have:
\[ \frac{x}{-\frac{2}{7}} = \frac{-\frac{7}{2}}{x} \]

3. **Cross-multiply:**
Cross-multiplying gives us:
\[ x^2 = -\frac{7}{2} \cdot -\frac{2}{7} \]

4. **Simplify the right side:**
Simplifying the right side:
\[ -\frac{7}{2} \cdot -\frac{2}{7} = \frac{7 \cdot 2}{2 \cdot 7} = 1 \]
So, we have:
\[ x^2 = 1 \]

5. **Solve for \( x \):**
Taking the square root of both sides gives us:
\[ x = \pm 1 \]

### Final Answer:
The values of \( x \) that make the numbers \( -\frac{2}{7}, x, -\frac{7}{2} \) in G.P. are:
\[ x = 1 \quad \text{or} \quad x = -1 \]