Welcome to Sequences & Series. A sequence is one of the simplest ideas in maths — just a list of numbers in a fixed order — but it shows up everywhere, and it is a heavy topic on the CSCA exam (about one question in five across the paper).
This is the first of five lessons. Here we set up the language: how we name the terms of a sequence, how to read off the terms when we are handed a formula, how to find the formula when we are handed a pattern, and what it means for a sequence to be built by a rule that makes each term from the one before. None of this is hard, but it is the groundwork the next four lessons stand on.
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What is a sequence?
A sequence is an ordered list of numbers. Each number in the list is a term. For example, 2, 4, 6, 8, 10, … is the sequence of even numbers, and 1, 4, 9, 16, … is the sequence of perfect squares. The little "…" at the end just means the list keeps going by the same pattern.
Order is everything, so we label the terms by their position. The 1st term is written a₁, the 2nd is a₂, the 3rd is a₃, and the term in position n is aₙ. The whole sequence — all the terms together — is written {aₙ}. Here n is simply the position: a counting number, 1, 2, 3, and so on.
Every term has a position. The term in position n is written aₙ — so here a₁ = 3, a₂ = 5, a₃ = 7, and so on.
Keep two things apart: aₙ (with a number or letter in the slot) is a single term — one number — while {aₙ} is the whole sequence. So a₃ = 7 means "the third term is 7," and {aₙ} refers to the entire list 3, 5, 7, 9, 11, ….
Sequence and term. A sequence is an ordered list of numbers. Each number is a term. The term in position n is aₙ; the first term is a₁; the whole sequence is {aₙ}. The position n is a counting number: 1, 2, 3, ….
Try it! · Check-in question
In the sequence 5, 8, 11, 14, 17, …, what is a₄?
a₄ means the term in position 4. Counting along — 5 (1st), 8 (2nd), 11 (3rd), 14 (4th) — gives a₄ = 14, so the answer is B. Option A (11) is a₃, and option C (17) is a₅. Option D (4) is the position n, not the term that sits there.
The general term: from a formula to the terms
Very often a sequence is handed to you as a formula for aₙ in terms of n — this is called the general term. To get the actual numbers, you just substitute n = 1, 2, 3, … one at a time.
Worked example · listing terms from a formula
A sequence has general term aₙ = 2n + 1. Write the first four terms, and find a₁₀.
So the sequence is −1, 2, −3, 4, …. The (−1)ⁿ factor is −1 when n is odd and +1 when n is even, which makes the signs flip back and forth.
General term. A formula for aₙ in terms of the position n. Substitute n = 1, 2, 3, … to get the terms; substitute any single value of n to get that one term.
The sign switch. To make signs alternate, multiply by a power of −1. (−1)ⁿ gives −, +, −, +, … (negative first). (−1)ⁿ⁺¹ gives +, −, +, −, … (positive first). Pick whichever matches the very first term.
Try it! · Check-in question
If aₙ = (−1)ⁿ(2n − 1), what is a₃?
Put n = 3: a₃ = (−1)³(2·3 − 1) = (−1)(5) = −5. The answer is C. Option B (5) forgets the sign — (−1)³ is −1, not +1. Options A (−6) and D (6) use 2n = 6 instead of 2n − 1 = 5.
Finding the general term from a pattern
The reverse task is just as common on the exam: you are shown the first few terms and asked for a formula for aₙ. The trick is to handle the size of the terms and their sign separately, then put the two together.
1
Look at the sizes, ignoring any minus signs. Match them to a familiar pattern: the counting numbers (n), the squares (n²), powers (2ⁿ, 3ⁿ, …), or fractions (1/n, 1/2ⁿ, …).
2
If the signs alternate, attach a power of −1: use (−1)ⁿ if the first term is negative, or (−1)ⁿ⁺¹ if the first term is positive.
3
Combine the size and the sign into one formula — then test it on n = 1, 2, 3 to be sure.
Worked example · a pattern of sizes
Find the general term of 1, 4, 9, 16, 25, ….
There are no minus signs, so we only need the size rule. The terms are 1, 4, 9, 16, 25 — these are 1², 2², 3², 4², 5².
So the nth term is aₙ = n². Check: n = 1 gives 1, n = 2 gives 4, n = 3 gives 9. It works.
Worked example · sizes and alternating signs (April 2026, Q29)
Find the general term of 1/2, −1/4, 1/8, −1/16, ….
Size first. Ignoring signs, the terms are 1/2, 1/4, 1/8, 1/16 — that is 1/2ⁿ (the denominators are 2¹, 2², 2³, 2⁴).
Now the sign. The first term is positive, then it alternates: +, −, +, −. That is the (−1)ⁿ⁺¹ pattern (positive first).
Always run the check in step 3. A formula can match the first term by luck and still be wrong — testing n = 1, 2, and 3 catches almost every slip, especially a sign that starts on the wrong foot.
Try it! · Check-in question
What is the general term of the sequence 1, −1/2, 1/4, −1/8, …?
Sizes: 1, 1/2, 1/4, 1/8 = 1/2ⁿ⁻¹ (so that n = 1 gives 1/2⁰ = 1). Signs: +, −, +, … starting positive, which is (−1)ⁿ⁻¹. Combining, aₙ = (−1)ⁿ⁻¹ / 2ⁿ⁻¹, the answer is D. Option B uses (−1)ⁿ, which starts negative. Option C copies the 1/2ⁿ size from the worked example, but this sequence starts at 1, so C gives 1/2 as the first term. Option A makes both slips. Testing n = 1 — which should give 1 — rules out A, B, and C at once.
Try it! · Check-in question
What is the general term of the sequence 2, 4, 8, 16, …? (official practice exam, Q19)
The terms 2, 4, 8, 16 are 2¹, 2², 2³, 2⁴ — powers of 2 — so aₙ = 2ⁿ. The answer is A. Option D (2ⁿ⁻¹) is off by one: it gives 1 at n = 1, not 2. Option B (2n) gives 2, 4, 6, 8 — it matches the first two terms but fails at n = 3 (2·3 = 6, not 8). Option C (n²) gives 1, 4, 9. This is exactly why step 3 — checking past the first couple of terms — matters.
Recursive sequences: building term by term
Some sequences are not given by a direct formula at all. Instead you get a starting term and a rule that tells you how to get each new term from the one before. This is a recursive definition. To find a particular term, you simply apply the rule over and over, one step at a time, until you reach it.
A recursive rule makes each term from the one before. Here a₁ = 1 and every step does "×2, then +1", giving 1, 3, 7, 15, 31, ….
Worked example · iterating a recursive rule
A sequence has a₁ = 1 and aₙ = 2aₙ₋₁ + 1 for n ≥ 2. Find a₅.
The rule aₙ = 2aₙ₋₁ + 1 says: to get a term, double the term before it and add 1. Apply it step by step:
So a₅ = 31. The only way to reach a₅ is to climb through a₂, a₃, a₄ first — there is no shortcut here.
Try it! · Check-in question
A sequence follows the rule aₙ₊₁ = 2aₙ + 1, and a₃ = 15. What is a₁?
This time we walk backwards. The rule says "double, then add 1", so to undo one step we "subtract 1, then halve". From a₃ = 15: a₂ = (15 − 1)/2 = 7. Then a₁ = (7 − 1)/2 = 3. Check forwards: 3 → 7 → 15. ✓ The answer is B. Option A (7) stops one step early; that is a₂. Option C (31) applies the rule forwards (2 · 15 + 1), which gives a₄. Option D (9/4) undoes the steps in the wrong order, halving first and then subtracting 1: 15/2 − 1 = 13/2, then 13/4 − 1 = 9/4.
Try it! · Check-in question
A sequence has a₁ = 1, a₂ = 3, and aₙ₊₂ = aₙ₊₁ + aₙ. What is a₅? (October 2025, Q40)
Here each term is the sum of the previous two. a₃ = a₂ + a₁ = 3 + 1 = 4; a₄ = a₃ + a₂ = 4 + 3 = 7; a₅ = a₄ + a₃ = 7 + 4 = 11. The answer is D. Option C (7) is a₄, one step short, and option B (18) is a₆, one step too far. Option A (8) adds the wrong pair, a₄ + a₁ = 7 + 1; the rule always adds the two terms just before.
That covers the essentials. Notice we have not yet given any special names to particular kinds of sequence — that comes next. In the following lessons we meet two famous families: sequences that grow by adding the same amount each time, and sequences that grow by multiplying by the same amount each time. Everything you have learned here — terms, the index n, general terms, and working step by step — carries straight over.
Your turn! · Practice questions
Practice · Question 1
A sequence has general term aₙ = n² + 1. Which term of the sequence is equal to 50?
Here the term is known and the position is not. Set aₙ = 50 and solve for n: n² + 1 = 50, so n² = 49 and n = 7 (n = −7 is impossible, because a position is a positive whole number). Check: a₇ = 49 + 1 = 50. ✓ The answer is C. Option A puts 50 in for n, which finds a₅₀ = 2501 instead. Option B stops at n² = 49 and forgets the square root. Option D looks for 50 among the squares and forgets the + 1: 50 is not a square, but 50 − 1 = 49 is.
Practice · Question 2
If aₙ = (−1)ⁿ⁺¹ · 2n, what is a₁ + a₂ + a₃ + a₄?
Write out the four terms. (−1)ⁿ⁺¹ is + when n is odd and − when n is even, so a₁ = 2, a₂ = −4, a₃ = 6, a₄ = −8. The sum is 2 − 4 + 6 − 8 = −4. (Pairing helps: (2 − 4) + (6 − 8) = −2 − 2.) The answer is A. Option B (20) ignores the signs: 2 + 4 + 6 + 8. Option C (4) uses (−1)ⁿ, which starts negative: −2 + 4 − 6 + 8. Option D (−8) is a₄ alone, not the sum.
Practice · Question 3
What is the general term of the sequence −1/5, 1/7, −1/9, 1/11, …? (January 2026, Q40)
Sizes: the denominators are 5, 7, 9, 11 = 2n + 3 (at n = 1, 2·1 + 3 = 5). Signs: −, +, −, + starting negative, which is (−1)ⁿ. So aₙ = (−1)ⁿ / (2n + 3), the answer is B. Option A uses (−1)ⁿ⁺¹, which starts positive. Option C takes the first denominator, 5, as the constant: 2n + 5 gives 7 at n = 1. Option D uses 2n + 1, the usual formula for odd numbers, without checking where they start: it gives 3 at n = 1, not 5.
Practice · Question 4
What is the general term of the sequence 5, 8, 11, 14, …?
Each term is 3 more than the one before, so the formula grows like 3n: try aₙ = 3n + c. At n = 1 we need 3 + c = 5, so c = 2 and aₙ = 3n + 2. Check: n = 2 gives 8, n = 3 gives 11. ✓ The answer is D. Option A (3n + 5) uses the first term as the constant, but then n = 1 gives 8, not 5. Option B (3ⁿ + 2) matches the first term (3 + 2 = 5) but gives 11 at n = 2, not 8. This is why step 3 tests more than one term. Option C (3n) has the right steps but starts at 3.
Practice · Question 5
A sequence has a₁ = 1 and aₙ₊₁ = aₙ + n for n ≥ 1. What is a₅?
Here the amount added changes, because the rule contains n. Use the n of the term you start from: a₂ = a₁ + 1 = 2; a₃ = a₂ + 2 = 4; a₄ = a₃ + 3 = 7; a₅ = a₄ + 4 = 11. The answer is C. Option A (15) adds the new position instead (2, 3, 4, 5): 1, 3, 6, 10, 15. Option B (7) stops one step early; that is a₄. Option D (5) treats n as a fixed 1 and adds 1 every time.
Things to remember
Concept
What to know
Sequence
An ordered list of numbers. The nth term is aₙ, the first term is a₁, and the whole sequence is {aₙ}.
The index n
The position — a counting number 1, 2, 3, …. aₙ is the term sitting in position n (not the number n itself).
General term
A formula for aₙ in terms of n. Substitute n = 1, 2, 3, … to get the terms.
Finding a term
Put the value of n into the general term (e.g. aₙ = n² + 1 → a₅ = 26). No need to list every earlier term.
Finding the rule
Match the sizes to a pattern (n, n², 2ⁿ, 1/2ⁿ, …), then fix the sign with (−1)ⁿ or (−1)ⁿ⁺¹.
