Learn: Sequences & Series

Review the key ideas, then practise whenever you’re ready.

The idea

A sequence is a list of numbers following a pattern, like $5, 8, 11, 14, \dots$ . A series is what you get when you add those numbers up ( $5 + 8 + 11 + 14$ ). Instead of writing out 50 numbers by hand, two jump rules let you teleport straight to any term or sum the whole list in one step.

The four essential formulas:

1
Arithmetic nth term - $a_n = a_1 + (n - 1)d$ - start at $a_1$ and add $(n - 1)$ jumps of size $d$ .
2
Arithmetic series sum - $S_n = \frac{n}{2}(a_1 + a_n)$ - add the first and last terms, then multiply by $\frac{n}{2}$ pairs.
3
Geometric nth term - $a_n = a_1 \cdot r^{n-1}$ - start at $a_1$ and multiply by the ratio $r$ a total of $(n - 1)$ times.
4
Geometric series sum - $S_n = \frac{a_1(r^n - 1)}{r - 1}$ - adds the first $n$ terms of a multiplying pattern in one step.

Quick memory tips

Fencepost rule: you start on the 1st term, so reaching the 10th term takes 9 jumps - always use $(n - 1)$
Same addition step = Arithmetic (add $d$ ); same multiplication ratio = Geometric (multiply by $r$ )
When first jumps grow steadily like $+3, +5, +7$ , the second difference is constant - a quadratic sequence!

Interactive Sequence & Series Visualizer

See the fencepost $(n-1)$ rule and Gauss’s rainbow pairing trick.

You already start standing on the 1st term ( $a_1 = 5$ )! So reaching the 5th term takes 4 jumps of +3, not 5:

a_5 = a_1 + (5 - 1)d = 5 + 4(3) = 17

Arithmetic (Add d)

nth term:

a_n = a_1 + (n-1)d

Sum of n terms:

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

Geometric (Multiply by r)

nth term:

a_n = a_1 \cdot r^{n-1}

Sum of n terms:

S_n = \frac{a_1(r^n - 1)}{r - 1}

Worked example

Every step explained - why we do it, and what happens.

1
$a_1 = 5, \quad d = 3, \quad n = 20$
Why: In the sequence 5, 8, 11, 14, …, each step adds +3. We want the 20th term (a₂₀) and the sum of the first 20 terms (S₂₀).
What happens: Start a₁ = 5, common difference d = 3, and term count n = 20.
2
$a_{20} = 5 + (20 - 1) \times 3$
Why: Starting on the 1st term, reaching the 20th term takes 20 − 1 = 19 jumps of +3.
What happens: 19 jumps × 3 = 57 added to the starting 5.
3
$a_{20} = 5 + 57 = 62$
The 20th term is 62.
4
$a_1 + a_{20} = 5 + 62 = 67$
Why: Use Gauss’s rainbow pairing trick: pair the 1st and 20th terms (5 + 62 = 67), the 2nd and 19th terms (8 + 59 = 67), and so on.
What happens: 20 terms form 10 pairs, and every pair adds to 67.
5
$S_{20} = \frac{20}{2}(5 + 62) = 10 \times 67 = 670$
Why: Multiply 10 pairs by 67 per pair.
What happens: The sum of the first 20 terms is 670 ✓.

Second Differences & Sigma Notation (Optional Toolbox)

Two extra tools for Level 5 & Mastery: spotting quadratic sequences and reading $\sum$ sums.

Levels 5–6

1. Second Differences (Quadratic)

In 2, 5, 10, 17, …, the first jumps are +3, +5, +7. Those jumps grow by +2 each time! So the next jump is +9, giving 17 + 9 = 26.

2. Sigma Notation ( $\Sigma$ )

$\sum_{k=1}^{5} (2k + 1)$ simply means: plug in $k = 1$ (first term 3) and $k = 5$ (last term 11), then average and multiply by 5: $\frac{5(3 + 11)}{2} = 35$ .

Real-world example

You start a 12-week coin jar challenge: you put 10 coins in the jar in Week 1, 14 coins in Week 2, 18 coins in Week 3, adding 4 more coins each week. How many coins do you put in during Week 12, and how many total coins are in the jar after 12 weeks?

1
$a_1 = 10, \quad d = 4, \quad n = 12$
Why: You start at 10 coins (a₁ = 10), add 4 coins each week (d = 4), for 12 weeks (n = 12).
What happens: An arithmetic sequence with 12 terms.
2
$a_{12} = 10 + 11 \times 4 = 54$
Why: From Week 1 to Week 12 is 11 weekly increases of 4 coins (10 + 44 = 54).
What happens: In Week 12 alone, you put 54 coins into the jar.
3
$\text{Trap: } a_{12} = 10 + 12 \times 4 = 58\,?$
Why: The glitch: multiplying 4 by 12 instead of 11. Week 1 already had 10 coins, so there are only 11 jumps after Week 1.
What happens: 58 would be Week 13 - Week 12 is 54 coins.
4
$S_{12} = \frac{12}{2}(10 + 54) = 6 \times 64 = 384$
Why: Pair the 12 weeks into 6 pairs. Each pair (Week 1 + Week 12) totals 10 + 54 = 64 coins.
What happens: 6 × 64 = 384 coins in the jar after 12 weeks ✓.

Common glitch

Three classic sequence & series glitches:

1.Using $n$ jumps instead of $(n - 1)$ jumps

Wrong: $a_{10} = a_1 + 10d$ or $a_n = a_1 \cdot r^n$ - that lands on the 11th term

Right:Term 1 is your starting spot, so always use $(n - 1)$ jumps: $a_n = a_1 + (n - 1)d$ and $a_n = a_1 \cdot r^{n-1}$ .

2.Multiplying $a_1 \times r$ before the exponent in a geometric sequence

Wrong:In $a_4 = 3 \times 2^3$ , multiplying $3 \times 2 = 6$ first and computing $6^3 = 216$

Right:Exponents come before multiplication: compute $2^3 = 8$ first, then $3 \times 8 = 24$ .

3.Using an addition rule on a geometric (multiplying) pattern

Wrong:Finding the next term of $2, 6, 18$ by adding $+4$ or $+12$

Right:Check the ratio: $6 \div 2 = 3$ and $18 \div 6 = 3$ . Multiply by $3$ to get $18 \times 3 = 54$ .

General Calculator - Evaluate nth terms, powers, and series sums quickly.
Interest Calculator - See geometric growth in action with compound interest over time.

Next up: Coordinate Geometry →

Explore the next topic.

Ready to practise? Start with a warm-up, or return to the ladder.

Start Warm-Up Level →Back to Ladder

Learning is optional - the ladder is always open.

Loading math...