Find the \( 8^{\text {th }} \) term of the geometric sequence whose common ratio is \( \frac{3}{2} \) and whose first term is 5 .

To find the \( n \)-th term of a geometric sequence, we can use the formula: \[ a_n = a_1 \cdot r^{n-1} \] where: - \( a_n \) is the \( n \)-th term, - \( a_1 \) is the first term, - \( r \) is the common ratio, - \( n \) is the term number. In this case, we have: - \( a_1 = 5 \) - \( r = \frac{3}{2} \) - \( n = 8 \) Now, we can substitute these values into the formula: \[ a_8 = 5 \cdot \left( \frac{3}{2} \right)^{8-1} \] Calculating \( 8-1 \): \[ 8-1 = 7 \] Now we can rewrite the equation: \[ a_8 = 5 \cdot \left( \frac{3}{2} \right)^7 \] Next, we need to calculate \( \left( \frac{3}{2} \right)^7 \): \[ \left( \frac{3}{2} \right)^7 = \frac{3^7}{2^7} \] Calculating \( 3^7 \) and \( 2^7 \): \[ 3^7 = 2187 \] \[ 2^7 = 128 \] Thus, \[ \left( \frac{3}{2} \right)^7 = \frac{2187}{128} \] Now substituting this back into the equation for \( a_8 \): \[ a_8 = 5 \cdot \frac{2187}{128} \] This simplifies to: \[ a_8 = \frac{5 \cdot 2187}{128} = \frac{10935}{128} \] Thus, the \( 8^{\text{th}} \) term of the geometric sequence is: \[ \boxed{\frac{10935}{128}} \]