Write a rule for the \( n \)th term of the sequence. Then graph the first six terms of the sequence. \( a_{13}=42, d=4 \) \( a_{n}= \)

1. The \( n \)th term for an arithmetic sequence is given by \[ a_n = a_1 + (n-1)d. \] 2. We know that \( a_{13} = 42 \) and \( d = 4 \). To find \( a_1 \), substitute \( n = 13 \): \[ 42 = a_1 + (13-1) \cdot 4 = a_1 + 12 \cdot 4 = a_1 + 48. \] Solving for \( a_1 \): \[ a_1 = 42 - 48 = -6. \] 3. Now substitute \( a_1 = -6 \) and \( d = 4 \) into the general formula: \[ a_n = -6 + (n-1) \cdot 4. \] Simplify the expression: \[ a_n = -6 + 4n - 4 = 4n - 10. \] Thus, the rule for the \( n \)th term is \[ a_n = 4n - 10. \] 4. Next, determine the first six terms of the sequence: - For \( n = 1 \): \[ a_1 = 4(1) - 10 = -6. \] - For \( n = 2 \): \[ a_2 = 4(2) - 10 = -2. \] - For \( n = 3 \): \[ a_3 = 4(3) - 10 = 2. \] - For \( n = 4 \): \[ a_4 = 4(4) - 10 = 6. \] - For \( n = 5 \): \[ a_5 = 4(5) - 10 = 10. \] - For \( n = 6 \): \[ a_6 = 4(6) - 10 = 14. \] 5. To graph the first six terms, plot the points: - \((1, -6)\) - \((2, -2)\) - \((3, 2)\) - \((4, 6)\) - \((5, 10)\) - \((6, 14)\) The graph below represents these points on a coordinate plane: ``` 16 | 14 | • (6,14) 12 | 10 | • (5,10) 8 | 6 | • (4,6) 4 | 2 | • (3,2) 0 | -2 | • (2,-2) -4 | -6 |• (1,-6) --------------------------- 1 2 3 4 5 6 ``` Thus, the rule for the \( n \)th term is \[ a_n = 4n - 10, \] and the points of the sequence are as shown above.