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

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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 $$ \frac{1}{324} $$.
The sixth term of the sequence is $$ \frac{1}{324} $$.

Which term of the sequence
has a value of

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