If the first 5 terms of a geometric sequence are \( \left\{12,6,3, \frac{3}{2}, \frac{3}{4}\right\} \) then the formula for the \( n^{\text {th }} \) term in the sequence is a.) \( a_{n}=12 \cdot 2^{n-1} \) b.) \( a_{n}=\left(\frac{1}{2}\right) \cdot 12^{n-1} \) c.) \( a_{n}=2 \cdot 12^{n-1} \) d.) \( a_{n}=12 \cdot\left(\frac{1}{2}\right)^{n-1} \) (

To determine the formula for the \( n^{\text{th}} \) term of the given geometric sequence \( \left\{12, 6, 3, \frac{3}{2}, \frac{3}{4}\right\} \), follow these steps: 1. **Identify the First Term (\( a_1 \)) and the Common Ratio (\( r \)):** - The first term \( a_1 \) is **12**. - To find the common ratio \( r \), divide the second term by the first term: \[ r = \frac{6}{12} = \frac{1}{2} \] - Verify the ratio with subsequent terms: \[ \frac{3}{6} = \frac{1}{2}, \quad \frac{\frac{3}{2}}{3} = \frac{1}{2}, \quad \frac{\frac{3}{4}}{\frac{3}{2}} = \frac{1}{2} \] The common ratio \( r \) is consistently \( \frac{1}{2} \). 2. **Use the Formula for the \( n^{\text{th}} \) Term of a Geometric Sequence:** The general formula for the \( n^{\text{th}} \) term of a geometric sequence is: \[ a_n = a_1 \cdot r^{n-1} \] Substituting the known values: \[ a_n = 12 \cdot \left(\frac{1}{2}\right)^{n-1} \] 3. **Compare with the Given Options:** - **Option d.)** \( a_n = 12 \cdot \left(\frac{1}{2}\right)^{n-1} \) matches our derived formula. **Answer:** d.) \( a_{n}=12 \cdot\left(\frac{1}{2}\right)^{n-1} \)