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

To determine which term of the sequence \( 32, -16, 8, \ldots \) is \( \frac{1}{8} \), let's analyze the sequence step by step. ### Step 1: Identify the Sequence Type The given sequence is: - 1st term: \( 32 \) - 2nd term: \( -16 \) - 3rd term: \( 8 \) We can observe that each term is obtained by multiplying the previous term by \(-\frac{1}{2}\): - \( 32 \times \left(-\frac{1}{2}\right) = -16 \) - \( -16 \times \left(-\frac{1}{2}\right) = 8 \) Thus, the sequence is a **geometric sequence** with: - **First term (\( a \))**: \( 32 \) - **Common ratio (\( r \))**: \( -\frac{1}{2} \) ### Step 2: Formula for the \( n \)-th Term The formula for the \( n \)-th term of a geometric sequence is: \[ a_n = a \cdot r^{n-1} \] We need to find \( n \) such that: \[ 32 \cdot \left(-\frac{1}{2}\right)^{n-1} = \frac{1}{8} \] ### Step 3: Solve for \( n \) First, express both sides with base 2: \[ 32 = 2^5 \quad \text{and} \quad \frac{1}{8} = 2^{-3} \] Substitute into the equation: \[ 2^5 \cdot \left(-\frac{1}{2}\right)^{n-1} = 2^{-3} \] Simplify the equation by expressing \(-\frac{1}{2}\) as \( (-1) \cdot 2^{-1} \): \[ 2^5 \cdot (-1)^{n-1} \cdot 2^{-(n-1)} = 2^{-3} \] Combine the exponents of 2: \[ 2^{5 - (n-1)} \cdot (-1)^{n-1} = 2^{-3} \] \[ 2^{6 - n} \cdot (-1)^{n-1} = 2^{-3} \] For the magnitudes to be equal and the expression to be positive: \[ 6 - n = -3 \quad \Rightarrow \quad n = 9 \] Additionally, since \( \frac{1}{8} \) is positive, \( (-1)^{n-1} \) must also be positive. This occurs when \( n-1 \) is even, which is true for \( n = 9 \). ### Final Answer The term \( \frac{1}{8} \) is the **9th term** of the sequence. **Answer:** 9