2. Which term of the sequence \( \frac{3}{4}: \frac{1}{4} ; \frac{1}{12} ; \) has a value of \( \frac{1}{324} \)

To determine which term of the sequence \( \frac{3}{4}, \frac{1}{4}, \frac{1}{12}, \ldots \) equals \( \frac{1}{324} \), let's analyze the pattern of the sequence. 1. **Identify the Pattern:** - **Term 1:** \( \frac{3}{4} \) - **Term 2:** \( \frac{1}{4} \) (obtained by dividing Term 1 by 3: \( \frac{3}{4} \div 3 = \frac{1}{4} \)) - **Term 3:** \( \frac{1}{12} \) (obtained by dividing Term 2 by 3: \( \frac{1}{4} \div 3 = \frac{1}{12} \)) - Following this pattern, each subsequent term is obtained by dividing the previous term by 3. 2. **General Formula:** The sequence is geometric with a common ratio of \( \frac{1}{3} \). The general term \( T_n \) can be expressed as: \[ T_n = \frac{3}{4} \times \left(\frac{1}{3}\right)^{n-1} \] 3. **Find the Term Equal to \( \frac{1}{324} \):** Set \( T_n = \frac{1}{324} \): \[ \frac{3}{4} \times \left(\frac{1}{3}\right)^{n-1} = \frac{1}{324} \] Divide both sides by \( \frac{3}{4} \): \[ \left(\frac{1}{3}\right)^{n-1} = \frac{4}{972} = \frac{1}{243} \] Since \( 243 = 3^5 \), we have: \[ \left(\frac{1}{3}\right)^{n-1} = \left(\frac{1}{3}\right)^5 \] Therefore: \[ n - 1 = 5 \implies n = 6 \] **Answer:** The sixth term of the sequence has a value of \( \frac{1}{324} \).