1. The problem asks for the 5th term of a harmonic sequence given the 2nd term is $\frac{1}{2}$ and the 6th term is $\frac{1}{5}$. 2. A harmonic sequence is a sequence whose terms are the reciprocals of an arithmetic sequence. If the harmonic sequence is $\{a_n\}$, then $b_n = \frac{1}{a_n}$ is an arithmetic sequence. 3. We know $a_2 = \frac{1}{2}$ and $a_6 = \frac{1}{5}$. Therefore $b_2 = 2$ and $b_6 = 5$. 4. Since $b_n$ is arithmetic, $b_n = b_1 + (n-1)d$ for some first term $b_1$ and common difference $d$. 5. Using the two known terms, set up equations: $$b_2 = b_1 + d = 2$$ $$b_6 = b_1 + 5d = 5$$ 6. Subtract the first from the second: $$b_6 - b_2 = 5 - 2 = 3 = 4d \implies d = \frac{3}{4}$$ 7. Substitute $d$ back to find $b_1$: $$b_1 + \frac{3}{4} = 2 \implies b_1 = 2 - \frac{3}{4} = \frac{8}{4} - \frac{3}{4} = \frac{5}{4}$$ 8. Now find $b_5$: $$b_5 = b_1 + 4d = \frac{5}{4} + 4 \times \frac{3}{4} = \frac{5}{4} + 3 = \frac{5}{4} + \frac{12}{4} = \frac{17}{4}$$ 9. Recall $a_5 = \frac{1}{b_5}$, so $$a_5 = \frac{1}{\frac{17}{4}} = \frac{4}{17}$$ **Final answer:** The 5th term of the harmonic sequence is $\boxed{\frac{4}{17}}$.