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

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: \[ a_n = 256 \left(\frac{1}{4}\right)^{n-1} \]