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

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UpStudy AI · Accepted Answer

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:**
   - The first term ($$ a_1 $$) is $$ \frac{3}{4} $$.
   - The second term ($$ a_2 $$) is $$ \frac{1}{4} $$, which is $$ \frac{1}{3} $$ of $$ a_1 $$.
   - The third term ($$ a_3 $$) is $$ \frac{1}{12} $$, which is $$ \frac{1}{3} $$ of $$ a_2 $$.
   
   This suggests that the sequence is geometric with a common ratio $$ r = \frac{1}{3} $$.

2. **General Formula for the $$ n $$-th Term:**
   $$
   a_n = a_1 \times r^{n-1} = \frac{3}{4} \times \left(\frac{1}{3}\right)^{n-1}
   $$

3. **Find $$ n $$ such that $$ a_n = \frac{1}{324} $$:**
   $$
   \frac{3}{4} \times \left(\frac{1}{3}\right)^{n-1} = \frac{1}{324}
   $$
   
   Solving for $$ n $$:
   $$
   \left(\frac{1}{3}\right)^{n-1} = \frac{4}{972} = \frac{1}{243}
   $$
   $$
   \left(\frac{1}{3}\right)^{n-1} = \left(\frac{1}{3}\right)^5
   $$
   $$
   n - 1 = 5 \implies n = 6
   $$

4. **Verification:**
   $$
   a_6 = \frac{3}{4} \times \left(\frac{1}{3}\right)^5 = \frac{3}{4} \times \frac{1}{243} = \frac{1}{324}
   $$

**Conclusion:** The 6th term of the sequence is $$ \frac{1}{324} $$.

**Answer:** The sixth term of the sequence is equal to 1⁄324.
The 6th term of the sequence is $$ \frac{1}{324} $$.

Which term of the sequence
has a value of ?

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