Arithmetic Sequences and Series

By the end of this lesson, you will be able to find the $n$th term of an arithmetic sequence, compute partial sums of arithmetic series, and apply these formulas to real-world problems.

The Big Idea

An arithmetic sequence is any list where consecutive terms differ by a constant (the common difference $d$). Every linear function restricted to integer inputs generates an arithmetic sequence — making arithmetic sequences the discrete version of linear functions.

Arithmetic Sequence

Definition: $a_n = a_1 + (n-1)d$

• $a_1$ = first term
• $d$ = common difference
• $n$ = term number

Alternative: $a_n = a_{n-1} + d$ (recursive form)

Identifying $d$: $d = a_2 - a_1 = a_3 - a_2$ (constant difference)

Arithmetic Series (Partial Sum)

The sum of the first $n$ terms:

$$S_n = \frac{n}{2}(a_1 + a_n) = \frac{n}{2}(2a_1 + (n-1)d)$$

Derivation: Gauss's method — write the sum forwards and backwards, add:

$S_n + S_n = n(a_1 + a_n) \implies S_n = \frac{n}{2}(a_1 + a_n)$

Example 1 — Finding the $n$th Term

Sequence: $4, 9, 14, 19, \ldots$. Find $a_{20}$.

$a_1 = 4$, $d = 5$

$a_{20} = 4 + 19(5) = 4 + 95 = 99$

Example 2 — Finding the Sum

Sum the arithmetic sequence $3 + 7 + 11 + \cdots + 99$.

$d = 4$; find $n$: $3 + (n-1)(4) = 99 \implies n = 25$

$S_{25} = \frac{25}{2}(3 + 99) = \frac{25}{2}(102) = 1275$

Example 3 — Finding the Common Difference

$a_1 = 8$ and $a_{10} = 35$. Find $d$ and $S_{10}$.

$35 = 8 + 9d \implies d = 3$

$S_{10} = \frac{10}{2}(8 + 35) = 5(43) = 215$

Example 4 — Application: Salary Growth

An employee starts at $42,000 and receives $2,500 raises annually. What is the total salary over 10 years?

$a_1 = 42000$, $d = 2500$, $a_{10} = 42000 + 9(2500) = 64500$

$S_{10} = \frac{10}{2}(42000 + 64500) = 5(106500) = $532,500$

Sigma Notation

$$\sum_{k=1}^{n} a_k = a_1 + a_2 + \cdots + a_n$$

For an arithmetic series: $\sum_{k=1}^{20} (3 + 2k) = \sum_{k=1}^{20} a_k$ where $a_1 = 5$, $d = 2$.