Question

part a
use the given arithmetic sequence to write an equation.
-11, -15, -19, -23, ...
a) $a_n = -11 + (n - 1)(4)$
b) $a_n = -11 + (n^2 - 1)(4)$
c) $a_n = -11 + (n + 1)(-4)$
d) $a_n = -11 + (n - 1)(-4)$
part b
find the 7th term of the sequence.
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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.
Here, \(a_1=-11\). To find \(d\), we calculate \(d=a_2 - a_1=-15-(-11)=-4\).

Step2: Substitute into Formula

Substitute \(a_1 = - 11\) and \(d=-4\) into the formula \(a_n=a_1+(n - 1)d\). We get \(a_n=-11+(n - 1)(-4)\).

Step1: Use the Formula from Part A

From Part A, we have the formula \(a_n=-11+(n - 1)(-4)\). We want to find the 7th term, so we substitute \(n = 7\) into the formula.

Step2: Calculate the 7th Term

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Answer:

D. \(a_n=-11+(n - 1)(-4)\)

Part B