1. The problem is to find the nth term formula for each linear sequence given. 2. The formula for the nth term of a linear sequence is generally $$a_n = dn + c$$ where $d$ is the common difference and $c$ is a constant. 3. To find $d$, subtract the first term from the second term. 4. To find $c$, substitute $n=1$ and the first term into the formula and solve for $c$. --- Sequence 1: 1, 4, 7, 10, 13, 16, ... - Common difference $d = 4 - 1 = 3$ - Using $a_1 = 1$, $$1 = 3(1) + c \Rightarrow c = 1 - 3 = -2$$ - Formula: $$a_n = 3n - 2$$ Sequence 2: 2, 8, 14, 20, 26, 32, ... - $d = 8 - 2 = 6$ - $2 = 6(1) + c \Rightarrow c = 2 - 6 = -4$ - Formula: $$a_n = 6n - 4$$ Sequence 3: 5, 12, 19, 26, 33, 40, ... - $d = 12 - 5 = 7$ - $5 = 7(1) + c \Rightarrow c = 5 - 7 = -2$ - Formula: $$a_n = 7n - 2$$ Sequence 4: 6, 13, 20, 27, 34, 41, ... - $d = 13 - 6 = 7$ - $6 = 7(1) + c \Rightarrow c = 6 - 7 = -1$ - Formula: $$a_n = 7n - 1$$ Sequence 5: 6, 14, 22, 30, 38, 46, ... - $d = 14 - 6 = 8$ - $6 = 8(1) + c \Rightarrow c = 6 - 8 = -2$ - Formula: $$a_n = 8n - 2$$ Sequence 6: 8, 16, 24, 32, 40, 48, ... - $d = 16 - 8 = 8$ - $8 = 8(1) + c \Rightarrow c = 8 - 8 = 0$ - Formula: $$a_n = 8n$$ Sequence 7: 10, 19, 28, 37, 46, 55, ... - $d = 19 - 10 = 9$ - $10 = 9(1) + c \Rightarrow c = 10 - 9 = 1$ - Formula: $$a_n = 9n + 1$$ Sequence 8: 9, 21, 33, 45, 57, 69, ... - $d = 21 - 9 = 12$ - $9 = 12(1) + c \Rightarrow c = 9 - 12 = -3$ - Formula: $$a_n = 12n - 3$$ Sequence 9: -11, 2, 15, 28, 41, 54, ... - $d = 2 - (-11) = 13$ - $-11 = 13(1) + c \Rightarrow c = -11 - 13 = -24$ - Formula: $$a_n = 13n - 24$$ Sequence 10: -13, -1, 11, 23, 35, 47, ... - $d = -1 - (-13) = 12$ - $-13 = 12(1) + c \Rightarrow c = -13 - 12 = -25$ - Formula: $$a_n = 12n - 25$$ Final answers: 1. $$a_n = 3n - 2$$ 2. $$a_n = 6n - 4$$ 3. $$a_n = 7n - 2$$ 4. $$a_n = 7n - 1$$ 5. $$a_n = 8n - 2$$ 6. $$a_n = 8n$$ 7. $$a_n = 9n + 1$$ 8. $$a_n = 12n - 3$$ 9. $$a_n = 13n - 24$$ 10. $$a_n = 12n - 25$$