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

To determine the rule for the \( n \)th term of the sequence, we start with the general formula for an arithmetic sequence: \[ a_n = a_1 + (n-1)d \] We know \( a_{13} = 42 \) and the common difference \( d = 4 \). First, we find \( a_1 \) using \( a_{13} \): \[ a_{13} = a_1 + (13-1) \cdot 4 \] \[ 42 = a_1 + 48 \] By solving for \( a_1 \): \[ a_1 = 42 - 48 = -6 \] Now, we have everything we need to write the formula for the \( n \)th term: \[ a_n = -6 + (n-1) \cdot 4 \] Simplifying the equation: \[ a_n = -6 + 4n - 4 = 4n - 10 \] Now, let's calculate and graph 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 \( n=6 \): \( a_6 = 4(6) - 10 = 14 \) The first six terms are: -6, -2, 2, 6, 10, 14. Now, let’s visualize that on a graph: Plot points: - (1, -6) - (2, -2) - (3, 2) - (4, 6) - (5, 10) - (6, 14) Graph these points and connect them with a straight line, since it's a linear sequence. The resulting graph will show an upward slope, demonstrating how the values increase steadily with each term. Fun fact: the slope of that line corresponds to the common difference, which is 4!