3. In a geometric sequence, \( t_{3}=20 \) and \( t_{6}=-540 \), find \( t_{n} \). ( 6 marks \( ) \)

To find the general term \( t_n \) of a geometric sequence where \( t_3 = 20 \) and \( t_6 = -540 \), follow these steps: 1. **Express the Terms Using the Geometric Sequence Formula** The \( n \)-th term of a geometric sequence is given by: \[ t_n = a \cdot r^{n-1} \] where: - \( a \) is the first term, - \( r \) is the common ratio. 2. **Set Up Equations Using the Given Terms** For \( t_3 = 20 \): \[ a \cdot r^{3-1} = a \cdot r^2 = 20 \quad \text{(Equation 1)} \] For \( t_6 = -540 \): \[ a \cdot r^{6-1} = a \cdot r^5 = -540 \quad \text{(Equation 2)} \] 3. **Find the Common Ratio \( r \)** Divide Equation 2 by Equation 1 to eliminate \( a \): \[ \frac{a \cdot r^5}{a \cdot r^2} = \frac{-540}{20} \implies r^3 = -27 \] \[ r = \sqrt[3]{-27} = -3 \] 4. **Find the First Term \( a \)** Substitute \( r = -3 \) into Equation 1: \[ a \cdot (-3)^2 = 20 \implies a \cdot 9 = 20 \implies a = \frac{20}{9} \] 5. **Write the General Term \( t_n \)** Substitute \( a = \frac{20}{9} \) and \( r = -3 \) into the general formula: \[ t_n = \frac{20}{9} \cdot (-3)^{n-1} \] **Final Answer:** \[ t_n = \frac{20}{9} \times (-3)^{\,n-1} \]