To solve this problem, we need to understand the relationship between the fundamental frequencies of the segments of a string and the original string.$\\$When a string is divided into segments, the fundamental frequency of each segment is inversely proportional to its length.$\\$Let's denote:$\\$• $L$ as the original length of the string.$\\$• $L_1, L_2, L_3$ as the lengths of the three segments.$\\$• $n$ as the original fundamental frequency of the string.$\\$• $n_1, n_2, n_3$ as the fundamental frequencies of the segments.$\\$The fundamental frequency of a string is given by:$\\ n = \frac{1}{2L} \sqrt{\frac{T}{\mu}} \\$where $T$ is the tension and $\mu$ is the linear mass density.$\\$For the segments, the fundamental frequencies are:$\\ n_1 = \frac{1}{2L_1} \sqrt{\frac{T}{\mu}}, \\ n_2 = \frac{1}{2L_2} \sqrt{\frac{T}{\mu}}, \\ n_3 = \frac{1}{2L_3} \sqrt{\frac{T}{\mu}}. \\$Since the tension and linear mass density are the same for all segments, we can write:$\\ n_1 = \frac{1}{2L_1}, \\ n_2 = \frac{1}{2L_2}, \\ n_3 = \frac{1}{2L_3}. \\$The original string is divided into three segments, so:$\\ L = L_1 + L_2 + L_3. \\$The original fundamental frequency is:$\\ n = \frac{1}{2L}. \\$Substituting the expressions for $L_1, L_2,$ and $L_3$ in terms of $n_1, n_2,$ and $n_3: \\ L_1 = \frac{1}{2n_1}, \\ L_2 = \frac{1}{2n_2}, \\ L_3 = \frac{1}{2n_3}. \\$Thus, the total length $L$ is:$\\ L = \frac{1}{2n_1} + \frac{1}{2n_2} + \frac{1}{2n_3}. \\$Substituting back into the expression for $n: \\ n = \frac{1}{2L} = \frac{1}{\frac{1}{2n_1} + \frac{1}{2n_2} + \frac{1}{2n_3}}. \\$Simplifying, we get:$\\ \frac{1}{n} = \frac{1}{n_1} + \frac{1}{n_2} + \frac{1}{n_3}. \\$Therefore, the correct option is Option 1.