An arithmetic series is the sum of an arithmetic sequence, often written with Sigma Notation. There is a shortcut that avoids adding term by term, and it comes from a neat pairing trick.
Key idea
$$S_n = \frac{n(a_1 + a_n)}{2} = \frac{n}{2}\big(2a_1 + (n - 1)d\big)$$ Add the first and last terms, multiply by the number of terms, and divide by 2.
Why it works
Write the sum forward and backward: $S = 1 + 2 + \cdots + 100$ and $S = 100 + 99 + \cdots + 1$. Add the two lines column by column: every column adds to 101, and there are 100 columns, so $2S = 100 \cdot 101$ and $S = 5050$. The same pairing works for any arithmetic sequence, because stepping forward by $d$ in one line and backward by $d$ in the other keeps each column's total equal to $a_1 + a_n$.
Worked examples
Worked example: First $n$ terms
Find the sum of the first 20 terms of $5, 8, 11, \ldots$
- $a_{20} = 5 + 19 \cdot 3 = 62$.
- $S_{20} = \frac{20(5 + 62)}{2} = 10 \cdot 67 = 670$.
Answer: $670$
Worked example: A range of numbers
Add the whole numbers from 15 to 60.
- Number of terms: $60 - 15 + 1 = 46$.
- $S = \frac{46(15 + 60)}{2} = 23 \cdot 75 = 1725$.
Answer: $1725$
Worked example: Seats in a theater
The first row has 20 seats and each row has 2 more than the last. There are 15 rows. How many seats?
- Last row: $20 + 14 \cdot 2 = 48$.
- Total: $\frac{15(20 + 48)}{2} = 510$.
Answer: 510 seats
Common mistake
Using the wrong number of terms, especially for a range like 15 to 60 (that is 46 numbers, not 45). Another slip is using $a_n = a_1 + nd$ for the last term; it should be $a_1 + (n - 1)d$, because the first term has had no steps added yet.