Mini-Lesson

Sequences and series

This mini-lesson covers AQA section D — Sequences and series: arithmetic and geometric sequences and series, sigma notation, the sum to infinity and its convergence condition, recurrence relations (increasing, decreasing, periodic), and the binomial expansion — including the negative and fractional index case with its validity condition.

Work through each screen, answer the questions as you go (a few are conceptual, most are calculations) and collect ⭐ stars. Everything here is A-level standard, drawn from the AQA 7357 subject content. Press Start when you are ready.

Sequences · arithmetic

Arithmetic sequences and series

An arithmetic sequence adds a fixed common difference d each time. With first term a:

u_n = a + (n − 1)d · S_n = n/2 [2a + (n − 1)d]equivalently S_n = n/2 (a + l), where l is the last term

Worked example

For 5, 8, 11, 14, …: a = 5, d = 3.

20th term: u₂₀ = 5 + 19 × 3 = 5 + 57 = 62. (It is n − 1, not n — the first term needs no jumps.)

Sum of 20 terms: S₂₀ = 20/2 [2(5) + 19(3)] = 10 × (10 + 57) = 10 × 67 = 670.

Check with the other formula: S₂₀ = 20/2 (5 + 62) = 10 × 67 = 670 ✓

How many terms? For 2, 5, 8, …, 59: solve 2 + 3(n − 1) = 59 ⇒ 3(n − 1) = 57 ⇒ n − 1 = 19 ⇒ n = 20. Never just divide the range by d.

Sequences · geometric

Geometric sequences and series

A geometric sequence multiplies by a fixed common ratio r each time.

u_n = a rⁿ⁻¹ · S_n = a(1 − rⁿ)/(1 − r)valid for r ≠ 1

For 2, 6, 18, 54, …: a = 2 and r = 3 (each term ÷ the one before). So u₅ = 2 × 3⁴ = 162, and S₅ = 2(1 − 3⁵)/(1 − 3) = 2(−242)/(−2) = 242.

Finding r: divide consecutive terms — r = u₂/u₁. If the ratio is not constant, the sequence is not geometric. And a negative r makes the terms alternate in sign.

Geometric growth turns up in every compound-interest and population question: a sum invested at 5% per year is a GP with r = 1.05.

Quick check

Evaluate the sigma

?Evaluate Σ_r=1⁵ (2r + 1).

Sequences · sum to infinity

The sum to infinity

If the terms of a geometric series shrink, the running total settles down to a finite limit. This happens exactly when the ratio is strictly between −1 and 1.

S_∞ = a/(1 − r), valid only when |r| < 1the series is then called convergent

Worked examples

18 + 12 + 8 + … : r = 12/18 = 2/3, and |2/3| < 1 ✓. So S_∞ = 18/(1 − 2/3) = 18/(1/3) = 54.

12 + 6 + 3 + … : r = 1/2. S_∞ = 12/(1 − 0.5) = 24.

If |r| ≥ 1 the series diverges and S_∞ simply does not exist. Writing a/(1 − r) for r = 2 gives a number, but it is meaningless — always state the condition.

Sort it

Arithmetic, geometric or neither?

Tap a sequence, then tap the family it belongs to. Look at the differences and the ratios.

➕ Arithmetic

✖️ Geometric

❓ Neither

Sequences · sigma

Sigma notation

Σ is shorthand for "add these up". The letter underneath is the counter, and the numbers tell you where to start and stop.

Σ_r=1⁵ (2r + 1) = 3 + 5 + 7 + 9 + 11 = 35substitute r = 1, 2, 3, 4, 5, then add

Useful facts:

Σ of a linear expression in r gives an arithmetic series — use S_n = n/2[2a + (n − 1)d].
Σ ar^k with a constant ratio gives a geometric series.
Σ_r=1ⁿ c = nc (adding a constant n times), and Σ (f(r) + g(r)) = Σf(r) + Σg(r).

Mind the limits. Σ_r=3¹⁰ has 8 terms, not 7 — count them: 10 − 3 + 1 = 8. Off-by-one here wrecks the whole answer.

Quick check

When does it converge?

?A geometric series has first term a and common ratio r. For what values of r does the sum to infinity exist?

Sequences · recurrence

Recurrence relations

A recurrence relation defines each term from the previous one, together with a starting value.

u_n+1 = 2u_n − 3, u₁ = 5⇒ 5, 7, 11, 19, 35, … (increasing)

AQA asks you to classify the behaviour:

Increasing: u_n+1 > u_n for all n.
Decreasing: u_n+1 < u_n for all n.
Periodic: the terms repeat in a cycle. u_n+1 = 3 − u_n with u₁ = 1 gives 1, 2, 1, 2, … — period 2.

Just generate the terms. Do not try to spot a closed form: write out five or six terms and the behaviour becomes obvious. Watch the subscripts — u_n+1 is the next term.

Calculate

Your turn — arithmetic term

1An arithmetic sequence has first term a = 5 and common difference d = 3. Find the 20th term.

Hint: u₂₀ = a + (n − 1)d = 5 + 19 × 3. It is 19 jumps, not 20 — the first term needs none.

Calculate

Your turn — arithmetic sum

2For the same sequence (a = 5, d = 3), find the sum of the first 20 terms.

Hint: S₂₀ = n/2[2a + (n − 1)d] = 10 × [10 + 57] = 10 × 67. Or S₂₀ = 20/2 × (first + last) = 10 × (5 + 62).

Series · binomial

The binomial expansion (positive integer n)

For a positive integer n the expansion of (a + b)ⁿ terminates after n + 1 terms:

(a + b)ⁿ = Σ ⁿC_r a^n−r b^rⁿC_r = n! / (r!(n − r)!) — the entries of Pascal's triangle

Worked example — pick out one coefficient

Find the coefficient of x³ in (1 + 2x)⁶.

The x³ term is ⁶C₃ × 1³ × (2x)³ = 20 × 8x³ = 160x³.

The coefficient is 160. Note the 2 gets cubed as well — forgetting to cube it gives 20, a classic slip.

Always bracket the whole term. (2x)³ = 8x³, not 2x³.

Quick check

State the validity

?For which values of x is the expansion of (1 − 3x)⁻² valid?

Calculate

Your turn — sum to infinity

3A geometric series has first term 18 and common ratio r = 2/3. Find its sum to infinity.

Hint: Since |2/3| < 1 the series converges. S∞ = a/(1 − r) = 18 ÷ (1 − 2/3) = 18 ÷ (1/3).

Series · binomial

Binomial with negative or fractional index

This is the A-level extension. For any real n the expansion is an infinite series:

(1 + x)ⁿ = 1 + nx + n(n−1)/2! x² + n(n−1)(n−2)/3! x³ + …valid only for |x| < 1

Worked example — (1 − 3x)⁻²

Here n = −2 and the bracket contains u = −3x.

x-term: n·u = (−2)(−3x) = 6x.

x²-term: n(n−1)/2 · u² = (−2)(−3)/2 × (−3x)² = 3 × 9x² = 27x².

So (1 − 3x)⁻² = 1 + 6x + 27x² + 108x³ + …

The validity condition uses the whole bracket: we need |−3x| < 1, i.e. |x| < 1/3. Quoting |x| < 1 here is wrong and is heavily penalised.

Match it

Match the question to its value

Tap an item on the left, then its partner on the right.

Question

Value

Series · using expansions

Using expansions: approximations and non-unit brackets

Two routine manipulations AQA expects:

Non-unit bracket: factor it out first. (4 + x)^1/2 = 4^1/2(1 + x/4)^1/2 = 2(1 + x/4)^1/2, valid for |x/4| < 1, i.e. |x| < 4.
Approximation: substitute a small x. Using (1 − 3x)⁻² ≈ 1 + 6x + 27x² at x = 0.01 gives 1 + 0.06 + 0.0027 = 1.0627; the true value 1/0.97² = 1.06275… — accurate to 4 d.p.

Partial fractions meet the binomial: to expand (5x + 1)/((x − 1)(x + 2)) you first split it into 2/(x − 1) + 3/(x + 2), then expand each bracket separately. The overall validity is the tighter of the two conditions.

Approximations only work because the terms shrink — which is exactly the |x| < 1 condition doing its job.

Quick check

Classify the sequence

?A sequence is defined by u_n+1 = 3 − u_n with u₁ = 1. How would you describe it?

Calculate

Your turn — binomial coefficient

4Find the coefficient of x³ in the expansion of (1 + 2x)⁶.

Hint: Term = ⁶C₃ × (2x)³ = 20 × 8x³. Remember to cube the 2 as well as the x.

Calculate

Your turn — negative index

5Expand (1 − 3x)⁻² in ascending powers of x. Find the coefficient of x².

Hint: With n = −2 and u = −3x, the x² term is n(n−1)/2! × u² = [(−2)(−3)/2] × (−3x)² = 3 × 9x². (Valid for |x| < 1/3.)

Recap

The big ideas to know

Arithmetic: uₙ = a + (n − 1)d · Sₙ = n/2[2a + (n − 1)d]

Geometric: uₙ = arⁿ⁻¹ · Sₙ = a(1 − rⁿ)/(1 − r) · S∞ = a/(1 − r) only if |r| < 1

Sigma: Σ from r = p to q has q − p + 1 terms

Recurrence: generate the terms — increasing, decreasing or periodic

Binomial (n ∈ ℕ): coefficient of x³ in (1 + 2x)⁶ is ⁶C₃ × 2³ = 160

Binomial (any n): infinite series, valid for |whole bracket term| < 1 — for (1 − 3x)⁻² that is |x| < 1/3

That is AQA section D — Sequences and series. Press Finish to see your score.

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