CERCISE 4 The third term of a geometric sequence is 6 and the \( \operatorname{sixth} \) term is \( \frac{3}{32} \). Determine the first three terms and the general term \( \mathrm{T}_{n} \).

Let's solve the problem step by step. **Given:** - The third term of a geometric sequence is \( a_3 = 6 \). - The sixth term of the sequence is \( a_6 = \frac{3}{32} \). **Objective:** - Determine the first three terms of the sequence. - Find the general term \( T_n \). ### Step 1: Understanding the Geometric Sequence In a geometric sequence, each term is obtained by multiplying the previous term by a common ratio \( r \). The \( n \)-th term of a geometric sequence can be expressed as: \[ a_n = a_1 \cdot r^{n-1} \] where: - \( a_1 \) is the first term. - \( r \) is the common ratio. - \( n \) is the term number. ### Step 2: Setting Up the Equations Given: \[ a_3 = a_1 \cdot r^{2} = 6 \quad \text{(1)} \] \[ a_6 = a_1 \cdot r^{5} = \frac{3}{32} \quad \text{(2)} \] ### Step 3: Solving for the Common Ratio \( r \) Divide equation (2) by equation (1) to eliminate \( a_1 \): \[ \frac{a_1 \cdot r^{5}}{a_1 \cdot r^{2}} = \frac{\frac{3}{32}}{6} \] \[ r^{3} = \frac{3}{32 \times 6} = \frac{3}{192} = \frac{1}{64} \] \[ r = \sqrt[3]{\frac{1}{64}} = \frac{1}{4} \] ### Step 4: Solving for the First Term \( a_1 \) Using equation (1): \[ a_1 \cdot \left(\frac{1}{4}\right)^{2} = 6 \] \[ a_1 \cdot \frac{1}{16} = 6 \] \[ a_1 = 6 \times 16 = 96 \] ### Step 5: Listing the First Three Terms Now that we have \( a_1 = 96 \) and \( r = \frac{1}{4} \), the first three terms are: \[ a_1 = 96 \] \[ a_2 = a_1 \cdot r = 96 \times \frac{1}{4} = 24 \] \[ a_3 = a_2 \cdot r = 24 \times \frac{1}{4} = 6 \] ### Step 6: Writing the General Term \( T_n \) Using the formula for the \( n \)-th term: \[ T_n = a_1 \cdot r^{n-1} = 96 \cdot \left(\frac{1}{4}\right)^{n-1} \] **Final Answer:** - **First three terms:** 96, 24, 6 - **General term:** \( T_n = 96 \times \left(\dfrac{1}{4}\right)^{\,n-1} \)