Question

this question has two parts. first, answer part a. then, answer part b.
part a
write an equation for the nth term of the arithmetic sequence.
7, 13, 19, 25, ...
a) $a_n = n + 1$
b) $a_n = 6n + 1$
c) $a_n = 5n + 1$
d) $a_n = 6n^2 + 1$
part b
graph the first five terms of the sequence.
a) graph of option a, with $a_n$ on the y - axis and some points plotted

Explanation:

Part A

Step1: Recall arithmetic sequence formula

The formula for the \( n \)-th term of an arithmetic sequence is \( a_n = a_1 + (n - 1)d \), where \( a_1 \) is the first term and \( d \) is the common difference.
For the sequence \( 7, 13, 19, 25, \dots \), \( a_1 = 7 \). The common difference \( d = 13 - 7 = 6 \).

Step2: Substitute into the formula

Substitute \( a_1 = 7 \) and \( d = 6 \) into \( a_n = a_1 + (n - 1)d \):
\[

$$\begin{align*} a_n&= 7 + (n - 1) \times 6\\ &= 7 + 6n - 6\\ &= 6n + 1 \end{align*}$$

\]
We can also check by plugging in \( n = 1, 2, 3, 4 \):

• For \( n = 1 \): \( a_1 = 6(1) + 1 = 7 \) (correct)
• For \( n = 2 \): \( a_2 = 6(2) + 1 = 13 \) (correct)
• For \( n = 3 \): \( a_3 = 6(3) + 1 = 19 \) (correct)
• For \( n = 4 \): \( a_4 = 6(4) + 1 = 25 \) (correct)

First, find the first five terms using the formula \( a_n = 6n + 1 \):

• When \( n = 1 \): \( a_1 = 6(1) + 1 = 7 \)
• When \( n = 2 \): \( a_2 = 6(2) + 1 = 13 \)
• When \( n = 3 \): \( a_3 = 6(3) + 1 = 19 \)
• When \( n = 4 \): \( a_4 = 6(4) + 1 = 25 \)
• When \( n = 5 \): \( a_5 = 6(5) + 1 = 31 \)

Now, check the graph options. The points should be \( (1, 7) \), \( (2, 13) \), \( (3, 19) \), \( (4, 25) \), \( (5, 31) \). Looking at the given graph option A, the y - axis ( \( a_n \)) values seem to match these terms (7, 13, 19, 25, 31) when plotted against \( n = 1, 2, 3, 4, 5 \).

Answer:

B) \( a_n = 6n + 1 \)

Part B