2.5 Fill in the missing terms/numbers below, a. \( 1 ; 3 ; 9 ; \ldots \ldots, \ldots \). b. \( -3 ; \ldots \ldots ; \ldots ; \ldots ; 2 ; \ldots \).

a. The sequence is geometric. Since each term is obtained by multiplying the previous one by the same number, we determine the common ratio: \[ r = \frac{3}{1} = 3. \] Thus, the missing terms can be found as follows: \[ 9 \times 3 = 27,\quad 27 \times 3 = 81,\quad 81 \times 3 = 243. \] So the complete sequence is: \[ 1 ;\; 3 ;\; 9 ;\; 27 ;\; 81 ;\; 243. \] b. The sequence appears to be arithmetic with the first term \(a_1 = -3\) and the fifth term \(a_5 = 2\). The general form for an arithmetic sequence is: \[ a_n = a_1 + (n-1)d. \] Since \(a_5 = -3 + 4d = 2\), we solve for \(d\): \[ -3 + 4d = 2 \quad \Longrightarrow \quad 4d = 5 \quad \Longrightarrow \quad d = \frac{5}{4}. \] Now, we can fill in the missing terms: \[ a_2 = -3 + \frac{5}{4} = -\frac{12}{4} + \frac{5}{4} = -\frac{7}{4}, \] \[ a_3 = -3 + 2\left(\frac{5}{4}\right) = -3 + \frac{10}{4} = -\frac{12}{4} + \frac{10}{4} = -\frac{1}{2}, \] \[ a_4 = -3 + 3\left(\frac{5}{4}\right) = -3 + \frac{15}{4} = -\frac{12}{4} + \frac{15}{4} = \frac{3}{4}, \] \[ a_6 = -3 + 5\left(\frac{5}{4}\right) = -3 + \frac{25}{4} = -\frac{12}{4} + \frac{25}{4} = \frac{13}{4}. \] So the complete sequence is: \[ -3 ;\; -\frac{7}{4} ;\; -\frac{1}{2} ;\; \frac{3}{4} ;\; 2 ;\; \frac{13}{4}. \]