A sequence is just a list of numbers marching in order — but a good list hides a rule. Find the rule, and you can leap straight to the hundredth term, or add a thousand numbers in a single line.
A list with a rule
Look at this list of numbers:
\[ 2,\ 5,\ 8,\ 11,\ 14,\ \dots \]It isn't random. Each number is three more than the one before it. That hidden instruction — "add 3 to get the next term" — is the rule, and a list of numbers that follows a rule is called a sequence. Each number in the list is a term, and we name them in order: the first term is \( a_1 \), the second is \( a_2 \), the third is \( a_3 \), and the term in slot \( n \) is \( a_n \).
Now suppose we don't just want to read the list — we want to add it up. The moment you start summing the terms, the sequence becomes a series:
\[ 2 + 5 + 8 + 11 + 14 + \dots \]That's the whole distinction in one sentence: a sequence is the list; a series is the running total. Everything in this lesson is about two questions — what is the rule that builds the list, and is there a shortcut for the total?
Sequence vs. series. Keep the two words straight: a sequence is the ordered list itself, like \( 2, 5, 8, 11 \). A series is what you get when you join those terms with plus signs and add them, like \( 2 + 5 + 8 + 11 \). One you read; the other you total.
Two families that run the show
Almost every sequence worth knowing in algebra belongs to one of two families, and they differ by a single word: do you add the same amount each step, or do you multiply by the same amount each step?
An arithmetic sequence adds a fixed number — the common difference \( d \) — at every step. Starting from \( 2 \) with \( d = 3 \):
\[ 2,\ 5,\ 8,\ 11,\ 14,\ \dots \qquad (\text{add } 3 \text{ each time}) \]A geometric sequence instead multiplies by a fixed number — the common ratio \( r \) — at every step. Starting from \( 2 \) with \( r = 3 \):
\[ 2,\ 6,\ 18,\ 54,\ 162,\ \dots \qquad (\text{multiply by } 3 \text{ each time}) \]Arithmetic sequences grow in steady, even strides — a staircase. Geometric sequences grow by repeated multiplication, so they start slow and then explode upward — a ramp that keeps getting steeper. Spotting which family you're in is the first move every single time: check whether the gap between terms is added (arithmetic) or whether the ratio between terms is multiplied (geometric).
The widget below lets you feel the difference. Toggle between arithmetic and geometric, then change the first term and the step. Watch the bars: in arithmetic mode they climb by equal heights, while in geometric mode each bar towers over the last. The readout below the bars shows the running sum — keep an eye on how much faster it races upward in geometric mode.
Tip To tell the families apart, test two things. Subtract consecutive terms: if you always get the same number, it's arithmetic and that number is \( d \). Divide consecutive terms: if you always get the same number, it's geometric and that number is \( r \). Only one of the two tests will come out constant.
Jumping to any term with a formula
Listing terms one by one is fine for the fifth term, but nobody wants to write out a hundred numbers to reach the hundredth. Here is the secret that makes that unnecessary: because the rule is the same at every step, you can count how many steps you've taken and apply the rule all at once.
For an arithmetic sequence, getting to the \( n \)th term means starting at \( a_1 \) and adding \( d \) a total of \( n - 1 \) times (the first term takes zero steps, the second takes one, and so on):
\[ a_n = a_1 + (n - 1)\,d \]For a geometric sequence, getting to the \( n \)th term means starting at \( a_1 \) and multiplying by \( r \) a total of \( n - 1 \) times:
\[ a_n = a_1 \cdot r^{\,n-1} \]The shape is identical — start at \( a_1 \), then apply the step \( n - 1 \) times. The only difference is whether "apply the step" means add \( d \) or multiply by \( r \). With our arithmetic list \( 2, 5, 8, \dots \), the tenth term is \( a_{10} = 2 + 9 \cdot 3 = 29 \) without writing a single term in between. With the geometric list \( 2, 6, 18, \dots \), the fifth term is \( a_5 = 2 \cdot 3^4 = 162 \). No listing required.
Why \( n - 1 \), not \( n \)? The first term is where you begin, before any step has been taken. So the second term is one step in, the third is two steps in, and the \( n \)th term is \( n - 1 \) steps in. That single off-by-one is the most common slip — always count steps from the first term, not from zero.
Adding it all up: the series shortcuts
Now for the totals. The sum of the first \( n \) terms is written \( S_n \), and each family has its own beautiful shortcut.
The arithmetic one comes from a trick a young boy named Gauss is said to have used to add \( 1 + 2 + 3 + \dots + 100 \) in seconds. Write the sum forwards, then write it again backwards underneath:
\[ \begin{aligned} S &= 1 + 2 + 3 + \dots + 100 \\ S &= 100 + 99 + 98 + \dots + 1 \end{aligned} \]Add the two rows column by column. Every column gives the same total, \( 1 + 100 = 101 \), and there are \( 100 \) columns, so \( 2S = 100 \cdot 101 \), which means \( S = 5050 \). The genius is that pairing the smallest with the largest, the next-smallest with the next-largest, and so on, makes every pair add to the same thing. That pairing is exactly the formula:
\[ S_n = \frac{n}{2}\,(a_1 + a_n) \]In words: the sum of an arithmetic series is the number of terms times the average of the first and last term. The geometric sum has its own clean shortcut (for any ratio \( r \neq 1 \)):
\[ S_n = \frac{a_1\,(r^{\,n} - 1)}{r - 1} \]You don't need to memorise where this one comes from to use it — just plug in the first term \( a_1 \), the ratio \( r \), and how many terms \( n \) you're adding. (When \( r = 1 \) every term is the same, so the sum is simply \( a_1 \cdot n \), and we use that instead.)
Tip For an arithmetic sum you need the last term \( a_n \). If you only know \( a_1 \), \( d \), and \( n \), find \( a_n = a_1 + (n-1)d \) first, then drop it into \( S_n = \tfrac{n}{2}(a_1 + a_n) \). The two formulas almost always work as a pair.
Putting it together
- Identify the family and the step. Consecutive terms differ by \( 7 - 3 = 4 \), so it's arithmetic with \( a_1 = 3 \) and \( d = 4 \).
- Find the 10th term with \( a_n = a_1 + (n-1)d \): \( a_{10} = 3 + (10 - 1)\cdot 4 = 3 + 36 = 39 \).
- Now sum with \( S_n = \tfrac{n}{2}(a_1 + a_n) \): \( S_{10} = \tfrac{10}{2}(3 + 39) = 5 \cdot 42 = 210 \).
So the 10th term is \( \mathbf{39} \) and the first ten terms add to \( \mathbf{210} \).
- Identify the family and the step. Each term is \( 6 \div 2 = 3 \) times the one before, so it's geometric with \( a_1 = 2 \) and \( r = 3 \).
- Find the 5th term with \( a_n = a_1 \cdot r^{\,n-1} \): \( a_5 = 2 \cdot 3^{\,5-1} = 2 \cdot 3^4 = 2 \cdot 81 = 162 \).
- Now sum with \( S_n = \dfrac{a_1(r^{\,n} - 1)}{r - 1} \): \( S_5 = \dfrac{2\,(3^5 - 1)}{3 - 1} = \dfrac{2 \cdot 242}{2} = 242 \).
So the 5th term is \( \mathbf{162} \) and the first five terms add to \( \mathbf{242} \).
Practice
Try each one yourself, then reveal the full solution.
1. The arithmetic sequence \( 5, 8, 11, \dots \) — find the 20th term, \( a_{20} \).
Consecutive terms differ by \( 8 - 5 = 3 \), so this is arithmetic with \( a_1 = 5 \) and \( d = 3 \).
Use \( a_n = a_1 + (n-1)d \) with \( n = 20 \): \( a_{20} = 5 + (20 - 1)\cdot 3 = 5 + 19 \cdot 3 \).
Compute: \( 5 + 57 = 62 \).
So \( a_{20} = \mathbf{62} \).
2. The geometric sequence \( 1, 2, 4, 8, \dots \) — find the 8th term, \( a_8 \).
Each term is \( 2 \div 1 = 2 \) times the previous one, so this is geometric with \( a_1 = 1 \) and \( r = 2 \).
Use \( a_n = a_1 \cdot r^{\,n-1} \) with \( n = 8 \): \( a_8 = 1 \cdot 2^{\,8-1} = 2^7 \).
Compute: \( 2^7 = 128 \).
So \( a_8 = \mathbf{128} \).
3. Add it up: \( 1 + 2 + 3 + \dots + 100 \) using the series formula.
This is an arithmetic series with \( a_1 = 1 \), last term \( a_n = 100 \), and \( n = 100 \) terms.
Use \( S_n = \tfrac{n}{2}(a_1 + a_n) \): \( S_{100} = \tfrac{100}{2}(1 + 100) = 50 \cdot 101 \).
Compute: \( 50 \cdot 101 = 5050 \).
So the total is \( \mathbf{5050} \) — exactly Gauss's answer, found in one line.