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

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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:**
$$
t_n = \frac{20}{9} \times (-3)^{\,n-1}
$$
The general term of the geometric sequence is $$ t_n = \frac{20}{9} \times (-3)^{n-1} $$.

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

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