To solve the problem, we need to establish the relationship between the frequency of the sonometer wire when it is vibrated in full length and the frequencies when it is divided into segments.
### Step-by-Step Solution:
1. **Understanding the Sonometer**:
A sonometer is an apparatus used to study the vibrations of strings. It consists of a wire stretched over a wooden board, with bridges that can divide the wire into segments.
2. **Frequency of Full Length**:
When the wire is vibrated in its full length \( L \), it has a frequency \( n \). The formula for the frequency of a vibrating string is given by:
\[
n = \frac{1}{2L} \sqrt{\frac{T}{m}}
\]
where \( T \) is the tension in the wire and \( m \) is the mass per unit length of the wire.
3. **Dividing the Wire**:
When the wire is divided into segments of lengths \( l_1, l_2, l_3, \ldots \), the frequencies of these segments will be \( n_1, n_2, n_3, \ldots \).
4. **Frequency of Each Segment**:
For each segment, the frequency can be expressed as:
\[
n_i = \frac{1}{2l_i} \sqrt{\frac{T}{m}}
\]
for \( i = 1, 2, 3, \ldots \)
5. **Relating Frequencies and Lengths**:
Since the tension \( T \) and mass per unit length \( m \) remain constant for the wire, we can relate the frequencies and lengths:
\[
n = \frac{1}{2L} \sqrt{\frac{T}{m}} \quad \text{and} \quad n_i = \frac{1}{2l_i} \sqrt{\frac{T}{m}}
\]
6. **Expressing Lengths in Terms of Frequencies**:
From the frequency formula, we can express the lengths in terms of frequencies:
\[
l_i = \frac{1}{2n_i} \sqrt{\frac{T}{m}}
\]
Therefore, the total length \( L \) can be expressed as:
\[
L = l_1 + l_2 + l_3 + \ldots
\]
7. **Substituting Lengths**:
Substituting the expressions for \( l_i \) into the equation for \( L \):
\[
L = \frac{1}{2n_1} \sqrt{\frac{T}{m}} + \frac{1}{2n_2} \sqrt{\frac{T}{m}} + \frac{1}{2n_3} \sqrt{\frac{T}{m}} + \ldots
\]
8. **Factoring Out Common Terms**:
Factoring out \( \sqrt{\frac{T}{m}} \):
\[
L = \sqrt{\frac{T}{m}} \left( \frac{1}{2n_1} + \frac{1}{2n_2} + \frac{1}{2n_3} + \ldots \right)
\]
9. **Relating Full Length Frequency to Segment Frequencies**:
Since \( n = \frac{1}{2L} \sqrt{\frac{T}{m}} \), we can relate this to the sum of the segment frequencies:
\[
\frac{1}{n} = \frac{1}{L} \cdot \sqrt{\frac{T}{m}} = \frac{1}{\sqrt{\frac{T}{m}}} \left( \frac{1}{2n_1} + \frac{1}{2n_2} + \frac{1}{2n_3} + \ldots \right)
\]
10. **Final Relationship**:
Thus, we arrive at the final relationship:
\[
\frac{1}{n} = \frac{1}{n_1} + \frac{1}{n_2} + \frac{1}{n_3} + \ldots
\]
### Final Answer:
The correct relation is:
\[
\frac{1}{n} = \frac{1}{n_1} + \frac{1}{n_2} + \frac{1}{n_3} + \ldots
\]
