1. **State the problem:** We have a geometric sequence with first term $a_1 = \frac{2}{3}$ and next three terms $\frac{1}{2}$, $\frac{3}{8}$, and $\frac{9}{32}$. We need to find the tenth term $a_{10}$. 2. **Formula for the $n$th term of a geometric sequence:** $$a_n = a_1 \cdot r^{n-1}$$ where $r$ is the common ratio. 3. **Find the common ratio $r$:** $$r = \frac{a_2}{a_1} = \frac{\frac{1}{2}}{\frac{2}{3}} = \frac{1}{2} \times \frac{3}{2} = \frac{3}{4}$$ 4. **Verify the ratio with other terms:** $$a_3 = a_2 \cdot r = \frac{1}{2} \times \frac{3}{4} = \frac{3}{8}$$ $$a_4 = a_3 \cdot r = \frac{3}{8} \times \frac{3}{4} = \frac{9}{32}$$ This confirms $r = \frac{3}{4}$. 5. **Calculate the tenth term:** $$a_{10} = a_1 \cdot r^{9} = \frac{2}{3} \times \left(\frac{3}{4}\right)^9$$ 6. **Simplify the power:** $$\left(\frac{3}{4}\right)^9 = \frac{3^9}{4^9} = \frac{19683}{262144}$$ 7. **Multiply:** $$a_{10} = \frac{2}{3} \times \frac{19683}{262144} = \frac{2 \times 19683}{3 \times 262144} = \frac{39366}{786432}$$ 8. **Simplify the fraction:** Divide numerator and denominator by 6: $$\frac{39366}{786432} = \frac{\cancel{6} 6561}{\cancel{6} 131072} = \frac{6561}{131072}$$ **Final answer:** $$a_{10} = \frac{6561}{131072}$$ This corresponds to option B.