Lesson 3.1.3: Arithmetic and Geometric Sequences - Algebra 1 With ... - Free Printable

To solve the problem, we need to analyze each sequence, identify its pattern, determine its type (arithmetic or geometric), and derive both the recursive and explicit formulas. Let's go through each sequence step by step.

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Sequence 1: \(-2, 2, 6, 10, \ldots\)

#### Step 1: Identify the Pattern
- The sequence is: \(-2, 2, 6, 10, \ldots\)
- Differences between consecutive terms:
\[
2 - (-2) = 4, \quad 6 - 2 = 4, \quad 10 - 6 = 4
\]
- The difference is constant (\(4\)).

#### Step 2: Type of Sequence
- Since the difference between consecutive terms is constant, this is an arithmetic sequence.

#### Step 3: Recursive Formula
- For an arithmetic sequence, the recursive formula is:
\[
A(n+1) = A(n) + d
\]
where \(d\) is the common difference.
- Here, \(d = 4\) and the first term \(A(1) = -2\).
- Therefore, the recursive formula is:
\[
A(n+1) = A(n) + 4, \quad A(1) = -2
\]

#### Step 4: Explicit Formula
- For an arithmetic sequence, the explicit formula is:
\[
A(n) = A(1) + (n-1)d
\]
- Substituting \(A(1) = -2\) and \(d = 4\):
\[
A(n) = -2 + (n-1) \cdot 4
\]
Simplify:
\[
A(n) = -2 + 4n - 4 = 4n - 6
\]

#### Final Answer for Sequence 1:
\[
\begin{aligned}
&\text{Pattern: add 4} \\
&\text{Type of Sequence: Arithmetic} \\
&\text{Recursive Formula: } A(n+1) = A(n) + 4, \quad A(1) = -2 \\
&\text{Explicit Formula: } A(n) = 4n - 6
\end{aligned}
\]

---

Sequence 2: \(2, 4, 8, 16, \ldots\)

#### Step 1: Identify the Pattern
- The sequence is: \(2, 4, 8, 16, \ldots\)
- Ratios between consecutive terms:
\[
\frac{4}{2} = 2, \quad \frac{8}{4} = 2, \quad \frac{16}{8} = 2
\]
- The ratio is constant (\(2\)).

#### Step 2: Type of Sequence
- Since the ratio between consecutive terms is constant, this is a geometric sequence.

#### Step 3: Recursive Formula
- For a geometric sequence, the recursive formula is:
\[
A(n+1) = A(n) \cdot r
\]
where \(r\) is the common ratio.
- Here, \(r = 2\) and the first term \(A(1) = 2\).
- Therefore, the recursive formula is:
\[
A(n+1) = A(n) \cdot 2, \quad A(1) = 2
\]

#### Step 4: Explicit Formula
- For a geometric sequence, the explicit formula is:
\[
A(n) = A(1) \cdot r^{n-1}
\]
- Substituting \(A(1) = 2\) and \(r = 2\):
\[
A(n) = 2 \cdot 2^{n-1}
\]
Simplify:
\[
A(n) = 2^n
\]

#### Final Answer for Sequence 2:
\[
\begin{aligned}
&\text{Pattern: multiply 2} \\
&\text{Type of Sequence: Geometric} \\
&\text{Recursive Formula: } A(n+1) = A(n) \cdot 2, \quad A(1) = 2 \\
&\text{Explicit Formula: } A(n) = 2^n
\end{aligned}
\]

---

Sequence 3: \(2, -3, -8, -13, \ldots\)

#### Step 1: Identify the Pattern
- The sequence is: \(2, -3, -8, -13, \ldots\)
- Differences between consecutive terms:
\[
-3 - 2 = -5, \quad -8 - (-3) = -5, \quad -13 - (-8) = -5
\]
- The difference is constant (\(-5\)).

#### Step 2: Type of Sequence
- Since the difference between consecutive terms is constant, this is an arithmetic sequence.

#### Step 3: Recursive Formula
- For an arithmetic sequence, the recursive formula is:
\[
A(n+1) = A(n) + d
\]
where \(d\) is the common difference.
- Here, \(d = -5\) and the first term \(A(1) = 2\).
- Therefore, the recursive formula is:
\[
A(n+1) = A(n) - 5, \quad A(1) = 2
\]

#### Step 4: Explicit Formula
- For an arithmetic sequence, the explicit formula is:
\[
A(n) = A(1) + (n-1)d
\]
- Substituting \(A(1) = 2\) and \(d = -5\):
\[
A(n) = 2 + (n-1) \cdot (-5)
\]
Simplify:
\[
A(n) = 2 - 5n + 5 = 7 - 5n
\]

#### Final Answer for Sequence 3:
\[
\begin{aligned}
&\text{Pattern: subtract 5} \\
&\text{Type of Sequence: Arithmetic} \\
&\text{Recursive Formula: } A(n+1) = A(n) - 5, \quad A(1) = 2 \\
&\text{Explicit Formula: } A(n) = 7 - 5n
\end{aligned}
\]

---

Final Table

\[
\begin{array}{|c|c|c|c|c|}
\hline
\text{Sequence} & \text{Pattern} & \text{Type of Sequence} & \text{Recursive Formula} & \text{Explicit Formula} \\
\hline
5, 8, 11, 14, \ldots & \text{add 3} & \text{Arithmetic} & A(n+1) = A(n) + 3, \quad A(1) = 5 & A(n) = 5 + 3(n-1) \text{ or } A(n) = 3n + 2 \\
\hline
5, \frac{5}{3}, \frac{5}{9}, \frac{5}{27}, \ldots & \text{multiply } \frac{1}{3} \text{ (divide 3)} & \text{Geometric} & A(n+1) = A(n) \cdot \frac{1}{3}, \quad A(1) = 5 & A(n) = 5 \cdot \left(\frac{1}{3}\right)^{n-1} \\
\hline
-2, 2, 6, 10, \ldots & \text{add 4} & \text{Arithmetic} & A(n+1) = A(n) + 4, \quad A(1) = -2 & A(n) = 4n - 6 \\
\hline
2, 4, 8, 16, \ldots & \text{multiply 2} & \text{Geometric} & A(n+1) = A(n) \cdot 2, \quad A(1) = 2 & A(n) = 2^n \\
\hline
2, -3, -8, -13, \ldots & \text{subtract 5} & \text{Arithmetic} & A(n+1) = A(n) - 5, \quad A(1) = 2 & A(n) = 7 - 5n \\
\hline
\end{array}
\]

Final Answer:

\[
\boxed{
\begin{array}{|c|c|c|c|c|}
\hline
\text{Sequence} & \text{Pattern} & \text{Type of Sequence} & \text{Recursive Formula} & \text{Explicit Formula} \\
\hline
5, 8, 11, 14, \ldots & \text{add 3} & \text{Arithmetic} & A(n+1) = A(n) + 3, \quad A(1) = 5 & A(n) = 5 + 3(n-1) \text{ or } A(n) = 3n + 2 \\
\hline
5, \frac{5}{3}, \frac{5}{9}, \frac{5}{27}, \ldots & \text{multiply } \frac{1}{3} \text{ (divide 3)} & \text{Geometric} & A(n+1) = A(n) \cdot \frac{1}{3}, \quad A(1) = 5 & A(n) = 5 \cdot \left(\frac{1}{3}\right)^{n-1} \\
\hline
-2, 2, 6, 10, \ldots & \text{add 4} & \text{Arithmetic} & A(n+1) = A(n) + 4, \quad A(1) = -2 & A(n) = 4n - 6 \\
\hline
2, 4, 8, 16, \ldots & \text{multiply 2} & \text{Geometric} & A(n+1) = A(n) \cdot 2, \quad A(1) = 2 & A(n) = 2^n \\
\hline
2, -3, -8, -13, \ldots & \text{subtract 5} & \text{Arithmetic} & A(n+1) = A(n) - 5, \quad A(1) = 2 & A(n) = 7 - 5n \\
\hline
\end{array}
}
\]