1 The first term of a geometric sequence is \( \frac{2}{3} \). The next three terms are \( \frac{1}{2}, \frac{3}{8} \), and \( \frac{9}{32} \). What is the tenth term of the sequence?

To find the tenth term of the geometric sequence, we first need to determine the common ratio \( r \). We can do this by dividing the second term by the first term: \[ r = \frac{\frac{1}{2}}{\frac{2}{3}} = \frac{1}{2} \cdot \frac{3}{2} = \frac{3}{4} \] Now that we have the common ratio \( r = \frac{3}{4} \), we can express the \( n \)-th term of a geometric sequence using the formula: \[ a_n = a_1 \cdot r^{n-1} \] where \( a_1 = \frac{2}{3} \) and \( r = \frac{3}{4} \). To find the tenth term \( a_{10} \): \[ a_{10} = a_1 \cdot r^{9} = \frac{2}{3} \cdot \left( \frac{3}{4} \right)^{9} \] Calculating \( \left( \frac{3}{4} \right)^{9} \): \[ \left( \frac{3}{4} \right)^{9} = \frac{3^9}{4^9} = \frac{19683}{262144} \] Now substituting back to find \( a_{10} \): \[ a_{10} = \frac{2}{3} \cdot \frac{19683}{262144} = \frac{2 \cdot 19683}{3 \cdot 262144} = \frac{39366}{786432} = \frac{39366 \div 39366}{786432 \div 39366} = \frac{1}{20} \approx \frac{1}{20} \] Thus, the tenth term of the sequence is \[ \frac{1}{20}. \]