Tell whether the sequence is arithmetic. Explain your reasoning. \( \frac{1}{6}, \frac{1}{2}, \frac{5}{6}, \frac{7}{6}, \frac{3}{2}, \ldots \) The sequence is arithmetic because the common difference is \( 1 / 6 \)

Let the terms of the sequence be: \[ a_1 = \frac{1}{6}, \quad a_2 = \frac{1}{2}, \quad a_3 = \frac{5}{6}, \quad a_4 = \frac{7}{6}, \quad a_5 = \frac{3}{2}, \ldots \] To determine if the sequence is arithmetic, we need to check whether the difference between consecutive terms is constant. 1. Calculate the difference between the second and first term: \[ a_2 - a_1 = \frac{1}{2} - \frac{1}{6} = \frac{3}{6} - \frac{1}{6} = \frac{2}{6} = \frac{1}{3} \] 2. Calculate the difference between the third and second term: \[ a_3 - a_2 = \frac{5}{6} - \frac{1}{2} = \frac{5}{6} - \frac{3}{6} = \frac{2}{6} = \frac{1}{3} \] 3. Calculate the difference between the fourth and third term: \[ a_4 - a_3 = \frac{7}{6} - \frac{5}{6} = \frac{2}{6} = \frac{1}{3} \] 4. Calculate the difference between the fifth and fourth term: \[ a_5 - a_4 = \frac{3}{2} - \frac{7}{6} = \frac{9}{6} - \frac{7}{6} = \frac{2}{6} = \frac{1}{3} \] Since the common difference is consistently \(\frac{1}{3}\) across the terms, the sequence is arithmetic. *Note:* The statement that the common difference is \(\frac{1}{6}\) is incorrect. The correct common difference is \(\frac{1}{3}\).