1. **State the problem:** We need to find the first four terms of a sequence starting at $\frac{5}{6}$ where each term is obtained by adding $\frac{2}{3}$ to the previous term. 2. **Formula and rule:** The sequence is arithmetic with first term $a_1 = \frac{5}{6}$ and common difference $d = \frac{2}{3}$. 3. **Calculate terms:** - Term 1: $a_1 = \frac{5}{6}$ - Term 2: $a_2 = a_1 + d = \frac{5}{6} + \frac{2}{3}$ Convert $\frac{2}{3}$ to sixths: $\frac{2}{3} = \frac{4}{6}$ So, $a_2 = \frac{5}{6} + \frac{4}{6} = \frac{9}{6} = \frac{\cancel{3} \times 3}{\cancel{3} \times 2} = \frac{3}{2}$ - Term 3: $a_3 = a_2 + d = \frac{3}{2} + \frac{2}{3}$ Find common denominator 6: $\frac{3}{2} = \frac{9}{6}$ and $\frac{2}{3} = \frac{4}{6}$ So, $a_3 = \frac{9}{6} + \frac{4}{6} = \frac{13}{6}$ - Term 4: $a_4 = a_3 + d = \frac{13}{6} + \frac{2}{3}$ Convert $\frac{2}{3} = \frac{4}{6}$ So, $a_4 = \frac{13}{6} + \frac{4}{6} = \frac{17}{6}$ 4. **Final answer:** The first four terms are: $$\frac{5}{6}, \frac{3}{2}, \frac{13}{6}, \frac{17}{6}$$ These match the given sequence values.