Write a rule for the $$ n $$th term of the sequence for which $$ a_{2}=64 $$ and $$ r=\frac{1}{4} $$. $$ a_{n}=256\left(\frac{1}{4}\right)^{n-1} $$

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To find the rule for the $$ n $$th term of the sequence given that $$ a_{2} = 64 $$ and $$ r = \frac{1}{4} $$, we can start by using the general formula for the $$ n $$th term of a geometric sequence:

$$
a_n = a_1 \cdot r^{n-1}
$$

Where:
- $$ a_n $$ is the $$ n $$th term,
- $$ a_1 $$ is the first term,
- $$ r $$ is the common ratio,
- $$ n $$ is the term number.

We know that $$ a_2 = 64 $$ and $$ r = \frac{1}{4} $$. We can express $$ a_2 $$ in terms of $$ a_1 $$:

$$
a_2 = a_1 \cdot r^{2-1} = a_1 \cdot r = a_1 \cdot \frac{1}{4}
$$

Setting this equal to 64 gives us:

$$
a_1 \cdot \frac{1}{4} = 64
$$

To find $$ a_1 $$, we can multiply both sides by 4:

$$
a_1 = 64 \cdot 4 = 256
$$

Now that we have $$ a_1 = 256 $$ and $$ r = \frac{1}{4} $$, we can write the rule for the $$ n $$th term:

$$
a_n = 256 \cdot \left(\frac{1}{4}\right)^{n-1}
$$

Thus, the rule for the $$ n $$th term of the sequence is:

$$
a_n = 256 \left(\frac{1}{4}\right)^{n-1}
$$
The $$ n $$th term of the sequence is $$ a_n = 256 \left(\frac{1}{4}\right)^{n-1} $$.

Write a rule for the \( n \)th term of the sequence for which \( a_{2}=64 \) and \( r=\frac{1}{4} \). \( a_{n}=256\left(\frac{1}{4}\right)^{n-1} \)

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