Which term of the sequence \( 32 ;-16 ; 8 ; \ldots \) is \( \frac{1}{8} ? \)

To identify which term in the sequence \( 32, -16, 8, \ldots \) equals \( \frac{1}{8} \), we first need to determine the pattern or rule of the sequence. This sequence appears to be a geometric sequence where each term is multiplied by a common ratio. The first term is 32, and the second term is -16. To find the common ratio, we divide the second term by the first term: \[ r = \frac{-16}{32} = -\frac{1}{2} \] Now we can express the general term of this sequence, denoted as \( a_n \), using the formula for the \( n \)-th term of a geometric sequence: \[ a_n = a_1 \cdot r^{(n-1)} = 32 \cdot \left(-\frac{1}{2}\right)^{(n-1)} \] We want to find \( n \) such that: \[ 32 \cdot \left(-\frac{1}{2}\right)^{(n-1)} = \frac{1}{8} \] To do so, we simplify the equation: \[ \left(-\frac{1}{2}\right)^{(n-1)} = \frac{1}{8 \cdot 32} = \frac{1}{256} \] Since \( 256 = 2^8 \), we can rewrite \( \frac{1}{256} \): \[ \left(-\frac{1}{2}\right)^{(n-1)} = \left(-\frac{1}{2}\right)^8 \] This means \( n-1 = 8 \). Therefore, solving for \( n \): \[ n = 9 \] Thus, the term of the sequence that is equal to \( \frac{1}{8} \) is the \( 9 \)-th term.