\( \frac { 1 } { 4 } , \frac { 6 } { 4 } , \frac { 9 } { 4 } \)

The sequence you provided consists of the fractions \( \frac{1}{4} \), \( \frac{6}{4} \), and \( \frac{9}{4} \). To analyze this sequence, we can look at the numerators: \( 1, 6, 9 \). 1. The first term is \( 1 \). 2. The second term is \( 6 \). 3. The third term is \( 9 \). The differences between the numerators are: - From \( 1 \) to \( 6 \): \( 6 - 1 = 5 \) - From \( 6 \) to \( 9 \): \( 9 - 6 = 3 \) The differences are not constant, so this does not appear to be an arithmetic sequence. However, if we look at the numerators more closely, we can see that: - \( 1 = 1^2 \) - \( 6 = 2^2 + 2 \) - \( 9 = 3^2 \) This suggests a possible pattern involving squares and some additional terms. If we want to find a general term for the sequence, we can express the numerators in terms of \( n \): - For \( n = 1 \): \( 1^2 = 1 \) - For \( n = 2 \): \( 2^2 + 2 = 6 \) - For \( n = 3 \): \( 3^2 = 9 \) To find the next term, we can look for a pattern. The next integer \( n = 4 \) gives us \( 4^2 = 16 \). Thus, the next term in the sequence would be: \[ \frac{16}{4} = 4 \] So the sequence continues as: \[ \frac{1}{4}, \frac{6}{4}, \frac{9}{4}, \frac{16}{4} \] If you have a specific question or need further analysis, please let me know!