AP Precalculus: Sequences
Master explicit and recursive formulas for arithmetic and geometric sequences
📚 Understanding Sequences
A sequence is an ordered list of numbers following a specific pattern. In AP Precalculus, you'll work with two main ways to define sequences: explicit formulas (direct calculation) and recursive formulas (building from previous terms). Mastering both approaches is essential for the exam.
1 Sequence Basics
A sequence is an ordered list of terms: \(a_1, a_2, a_3, \ldots, a_n, \ldots\). Each term is identified by its position number \(n\).
\(2, 5, 8, 11, 14, \ldots\)
\(a_1 = 2\), \(a_2 = 5\), \(a_3 = 8\), \(a_4 = 11\), \(a_5 = 14\)
2 Explicit vs. Recursive Formulas
There are two ways to define a sequence: explicitly (direct formula for any term) or recursively (each term defined using previous terms).
Explicit Advantage
Find any term directly without calculating all previous terms
Recursive Advantage
Shows the pattern of how each term relates to the previous
3 Arithmetic Sequences
An arithmetic sequence has a constant difference \(d\) between consecutive terms. Each term is found by adding \(d\) to the previous term.
Sequence: \(3, 7, 11, 15, 19, \ldots\)
Common difference: \(d = 7 - 3 = 4\)
Explicit: \(a_n = 3 + (n-1)(4) = 3 + 4n - 4 = 4n - 1\)
Recursive: \(a_1 = 3\), \(a_n = a_{n-1} + 4\)
Find \(a_{10}\): \(a_{10} = 4(10) - 1 = 39\)
4 Geometric Sequences
A geometric sequence has a constant ratio \(r\) between consecutive terms. Each term is found by multiplying the previous term by \(r\).
Sequence: \(2, 6, 18, 54, 162, \ldots\)
Common ratio: \(r = \frac{6}{2} = 3\)
Explicit: \(a_n = 2 \cdot 3^{n-1}\)
Recursive: \(a_1 = 2\), \(a_n = 3 \cdot a_{n-1}\)
Find \(a_6\): \(a_6 = 2 \cdot 3^5 = 2 \cdot 243 = 486\)
The exponent is \((n-1)\), not \(n\). For the first term, \(a_1 = a_1 \cdot r^0 = a_1 \cdot 1 = a_1\).
5 Identifying Sequence Type
Identify the type by checking whether consecutive terms have a constant difference (arithmetic) or constant ratio (geometric).
| Feature | Arithmetic | Geometric |
|---|---|---|
| Pattern | Add constant \(d\) | Multiply by constant \(r\) |
| Test | \(a_n - a_{n-1} =\) constant | \(\frac{a_n}{a_{n-1}} =\) constant |
| Explicit | \(a_n = a_1 + (n-1)d\) | \(a_n = a_1 \cdot r^{n-1}\) |
| Recursive | \(a_n = a_{n-1} + d\) | \(a_n = r \cdot a_{n-1}\) |
| Graph Shape | Linear | Exponential |
If you subtract consecutive terms and get the same value → Arithmetic. If you divide consecutive terms and get the same value → Geometric.
6 Converting Recursive ↔ Explicit
You can convert between recursive and explicit forms once you identify the sequence type and its parameters.
Recursive → Explicit
- Arithmetic: \(a_n = a_{n-1} + d\) → \(a_n = a_1 + (n-1)d\)
- Geometric: \(a_n = r \cdot a_{n-1}\) → \(a_n = a_1 \cdot r^{n-1}\)
- Identify \(a_1\), \(d\) or \(r\) from the recursive definition
Explicit → Recursive
- Arithmetic: Identify \(d\) from the explicit formula, write \(a_n = a_{n-1} + d\)
- Geometric: Identify \(r\) from the explicit formula, write \(a_n = r \cdot a_{n-1}\)
- Always include the initial term \(a_1\)
Given explicit: \(a_n = 5 + 3(n-1)\)
Identify: \(a_1 = 5\), \(d = 3\) (arithmetic)
Recursive: \(a_1 = 5\), \(a_n = a_{n-1} + 3\)
7 Finding Specific Terms
Use the appropriate formula to find any term in the sequence.
Using Explicit Formula
Substitute \(n\) directly into the formula to find \(a_n\)
Using Recursive Formula
Start with \(a_1\) and apply the rule repeatedly until reaching \(a_n\)
Given: \(a_1 = 4\), \(a_n = a_{n-1} + 7\)
Method 1 (Recursive): \(4 → 11 → 18 → 25 → 32 → 39 → 46 → 53\)
Method 2 (Convert to Explicit): \(a_n = 4 + (n-1)(7)\)
\(a_8 = 4 + (7)(7) = 4 + 49 = 53\)
For large \(n\) values, explicit is faster. For understanding pattern relationships, recursive is clearer.
📋 Quick Reference
Arithmetic Explicit
\(a_n = a_1 + (n-1)d\)
Arithmetic Recursive
\(a_n = a_{n-1} + d\)
Geometric Explicit
\(a_n = a_1 \cdot r^{n-1}\)
Geometric Recursive
\(a_n = r \cdot a_{n-1}\)
Common Difference
\(d = a_n - a_{n-1}\)
Common Ratio
\(r = \frac{a_n}{a_{n-1}}\)