Sequences Reference
Sequences and Series
A sequence lists terms in order. A series adds those terms. Most precalculus problems ask us to recognize whether the pattern grows by repeated addition or repeated multiplication.
Fact Table
| Pattern | Signal | Use |
|---|---|---|
| Arithmetic sequence | Constant difference $d$. | Linear explicit formula. |
| Geometric sequence | Constant ratio $r$. | Exponential explicit formula. |
| Finite arithmetic series | Add arithmetic terms. | Average first and last term, then multiply by count. |
| Finite geometric series | Add geometric terms. | Use the finite geometric sum formula. |
| Infinite geometric series | Geometric series with $|r|<1$. | The sum converges to a finite value. |
| Sigma notation | $\sum$ with an index. | Substitute each integer index value and add. |
Content Formulas
Arithmetic Sequence
$$a_n=a_1+(n-1)d$$
Geometric Sequence
$$a_n=a_1r^{n-1}$$
Arithmetic Series
$$S_n=\frac n2(a_1+a_n)$$
Finite Geometric Series
$$S_n=a_1\frac{1-r^n}{1-r},\quad r\ne1$$
Infinite Geometric Series
$$S_\infty=\frac{a_1}{1-r},\quad |r|<1$$
Recursive Form
$$a_n=a_{n-1}+d$$ $$a_n=ra_{n-1}$$
Check the pattern before choosing a formula. Arithmetic means add the same amount; geometric means multiply by the same amount.
Classic Examples
Arithmetic Term
Find the 20th term of $7, 11, 15, 19,\ldots$.
Solution StrategyThe common difference is $4$, so use the arithmetic explicit formula.
$$a_1=7,\quad d=4$$
$$a_{20}=7+(20-1)4$$
$$a_{20}=83$$
Geometric Term
Find the 8th term of $3, 6, 12, 24,\ldots$.
Solution StrategyThe common ratio is $2$, so use the geometric explicit formula.
$$a_1=3,\quad r=2$$
$$a_8=3(2)^{8-1}$$
$$a_8=384$$
Arithmetic Series
Find the sum of the first 30 terms of $5, 9, 13, 17,\ldots$.
Solution StrategyFind the last term, then average the first and last terms and multiply by the number of terms.
$$a_{30}=5+(30-1)4=121$$
$$S_{30}=\frac{30}{2}(5+121)$$
$$S_{30}=1890$$
Infinite Geometric Series
Find the sum $12+6+3+\frac32+\cdots$.
Solution StrategyThe common ratio is $\frac12$, so the infinite series converges.
$$a_1=12,\quad r=\frac12$$
$$S_\infty=\frac{12}{1-\frac12}$$
$$S_\infty=24$$