If `G_(1),G_(2)` are the geometric means fo two series of observations and G is the GM of the ratios of the corresponding observations then G is equal to

Options

A. `(G_(1))/(G_(2))`
B. `log G_(1) -log G_(2)`
C. `(log G_(1))/(log G_(2))`
D. `log(G_(1) *G_(2))`

Answer

To solve the problem, we need to establish the relationship between the geometric means of two series of observations and the geometric mean of the ratios of the corresponding observations.

### Step-by-Step Solution:

1. **Understanding Geometric Mean**:
The geometric mean (GM) of a set of numbers \( x_1, x_2, \ldots, x_n \) is given by:
\[ G = \sqrt[n]{x_1 \cdot x_2 \cdot \ldots \cdot x_n} \]

2. **Geometric Means of Two Series**:
Let’s denote the two series of observations as \( A = \{a_1, a_2, \ldots, a_n\} \) and \( B = \{b_1, b_2, \ldots, b_n\} \).
- The geometric mean of series \( A \) is:
\[ G_1 = \sqrt[n]{a_1 \cdot a_2 \cdot \ldots \cdot a_n} \]
- The geometric mean of series \( B \) is:
\[ G_2 = \sqrt[n]{b_1 \cdot b_2 \cdot \ldots \cdot b_n} \]

3. **Geometric Mean of Ratios**:
We need to find the geometric mean \( G \) of the ratios of corresponding observations, which is:
\[ G = \sqrt[n]{\frac{a_1}{b_1} \cdot \frac{a_2}{b_2} \cdots \frac{a_n}{b_n}} \]

4. **Simplifying the Expression**:
The expression for \( G \) can be rewritten as:
\[ G = \sqrt[n]{\frac{a_1 \cdot a_2 \cdots a_n}{b_1 \cdot b_2 \cdots b_n}} = \frac{\sqrt[n]{a_1 \cdot a_2 \cdots a_n}}{\sqrt[n]{b_1 \cdot b_2 \cdots b_n}} = \frac{G_1}{G_2} \]

5. **Final Result**:
Thus, we conclude that:
\[ G = \frac{G_1}{G_2} \]

### Conclusion:
The geometric mean \( G \) of the ratios of the corresponding observations is equal to the ratio of the geometric means \( G_1 \) and \( G_2 \) of the two series of observations.