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Arithmetic and geometric sequences and series: QCE Math Methods Unit 1 Year 11

Syllabus dot point

“Arithmetic and geometric sequences and series, including the general term formulas, sum formulas, and applications to growth and decay problems”

8 min read

Quick answer

Arithmetic sequences have constant common difference dd with Tn=a+(n−1)dT_n = a + (n-1)d and Sn=(n/2)[2a+(n−1)d]S_n = (n/2)[2a + (n-1)d]; geometric sequences have constant common ratio rr with Tn=arn−1T_n = ar^{n-1} and Sn=a(rn−1)/(r−1)S_n = a(r^n - 1)/(r-1), plus the infinite series S∞=a/(1−r)S_\infty = a/(1-r) for ∣r∣<1|r| < 1; both apply to compound interest, exponential growth and decay.

Jump to a section
  1. What this dot point is asking
  2. What a sequence is
  3. Arithmetic sequences
  4. Geometric sequences
  5. Recurrence and explicit forms
  6. Applications
  7. Sequence versus series
  8. Exam-style questions
Key fact

Previous study design. Arithmetic and geometric sequences and series (progressions) are not in the 2025 Mathematical Methods syllabus. This page follows the previous syllabus. It stays online for students using older resources. For the current course, start at the Mathematical Methods hub.

What this dot point is asking

QCAA wants Year 11 students to define arithmetic and geometric sequences and series, compute terms and sums, and apply them to growth and decay problems. This builds the algebraic fluency Year 12 Methods relies on, and connects directly to compound interest and exponential models.

What a sequence is

A sequence is an ordered list of numbers generated by a rule, with TnT_n denoting the nnth term. The two families studied here are defined by how each term relates to the previous one: arithmetic sequences add a fixed amount, geometric sequences multiply by a fixed factor. Identifying which family a sequence belongs to, by checking whether consecutive differences or consecutive ratios are constant, is the first decision in every problem and determines which formulas apply.

Arithmetic sequences

A sequence where each term differs from the previous by a constant dd (common difference).

General term. Tn=a+(n−1)dT_n = a + (n - 1) d where aa is the first term.

Sum of first nn terms. Sn=(n/2)[2a+(n−1)d]S_n = (n/2)[2a + (n - 1) d] or equivalently Sn=(n/2)(a+Tn)S_n = (n/2)(a + T_n).

Geometric sequences

A sequence where each term is the previous multiplied by a constant rr (common ratio).

General term. Tn=arn−1T_n = a r^{n-1}.

Sum of first nn terms. Sn=a(rn−1)/(r−1)S_n = a(r^n - 1)/(r - 1) for r≠1r \neq 1. (If r=1r = 1, Sn=naS_n = na.)

Infinite geometric series (for ∣r∣<1|r| < 1): S∞=a/(1−r)S_\infty = a / (1 - r).

The infinite formula requires convergence (∣r∣<1|r| < 1); otherwise the series diverges.

Recurrence and explicit forms

A sequence can be described either recursively, giving each term in terms of the previous one (for example Tn+1=Tn+dT_{n+1} = T_n + d for arithmetic, or Tn+1=rTnT_{n+1} = rT_n for geometric), or explicitly, giving TnT_n directly as a formula in nn. The explicit form is what you need to find a distant term without listing all the earlier ones, and deriving the explicit form from a described recurrence is a common first step in modelling questions.

Applications

Compound interest
A principal PP at rate rr per period compounded for nn periods grows to A=P(1+r)nA = P(1 + r)^n, and the successive balances form a geometric sequence with common ratio 1+r1 + r. This is the most direct application of geometric sequences in the course.
Exponential growth and decay
Population growth, radioactive decay and drug clearance all change by a constant factor per period, making them geometric and linking directly to the exponential functions studied alongside.
Annuities and repeated payments
A stream of regular payments earning compound interest sums as a geometric series, so the sum formula Sn=arn−1r−1S_n = a\dfrac{r^n - 1}{r - 1} values the whole stream. This idea underpins savings plans and loan repayments met later in the course.

Sequence versus series

A sequence is an ordered list of terms (T1,T2,T3,…T_1, T_2, T_3, \ldots); a series is the sum of its terms (Sn=T1+T2+⋯+TnS_n = T_1 + T_2 + \cdots + T_n). Questions often hinge on this distinction: "the fifth term" is T5T_5, a single value, while "the sum of the first five terms" is S5S_5. The general-term formulas describe the sequence, and the sum formulas describe the series. Reading which one a question wants is the first step to avoiding an avoidable error.

An arithmetic term and sum

Apply the term formula

A sequence has first term 33 and common difference 44. The tenth term is

T10=a+(n−1)d=3+9×4=39.T_{10} = a + (n - 1)d = 3 + 9 \times 4 = 39.

Apply the sum formula

S10=n2(a+T10)=102(3+39)=5×42=210.S_{10} = \frac{n}{2}(a + T_{10}) = \frac{10}{2}(3 + 39) = 5 \times 42 = 210.

So the first ten terms sum to 210210.

An infinite geometric sum

A geometric series has first term a=2a = 2 and common ratio r=0.5r = 0.5. Find the sum of all its terms.

Step 1: Check that the series converges

The sum-to-infinity formula only exists when each successive term becomes smaller in size, which happens precisely when ∣r∣<1|r| < 1. Here ∣r∣=0.5<1|r| = 0.5 < 1, so the terms shrink to zero and the series converges to a finite value. If we had ∣r∣≥1|r| \geq 1, the partial sums would grow without bound and no finite answer would exist.

Step 2: Apply the sum-to-infinity formula

S∞=a1−r=21−0.5=20.5=4.S_\infty = \frac{a}{1 - r} = \frac{2}{1 - 0.5} = \frac{2}{0.5} = 4.

Final answer: S∞=4S_\infty = 4.

Common errors
Off-by-one in TnT_n
Use n−1n - 1 in the exponent or multiplier, not nn.
Applying infinite formula when divergent
∣r∣<1|r| < 1 required.
Sign errors in (r−1)(r - 1) vs (1−r)(1 - r)
Use the form that matches; both are correct but make sure your sign convention is consistent.

Exam-style questions

Questions in the style of QCAA exam questions on this dot point, each with a worked answer. They are written by ExamExplained unless tagged "Past paper"; the year shows the paper a question is modelled on.

Practice4 marks
Paper 1 (technique). A geometric sequence has first term a=5a = 5 and common ratio r=2r = 2. (a) Determine the 88th term. (b) Determine the sum of the first 88 terms.
Show worked answer →

(a) Tn=arn−1T_n = ar^{n-1}, so T8=5×27=5×128=640T_8 = 5 \times 2^7 = 5 \times 128 = 640.

(b) Sn=arn−1r−1=528−12−1=5×255=1275S_n = a\dfrac{r^n - 1}{r - 1} = 5\dfrac{2^8 - 1}{2 - 1} = 5 \times 255 = 1275.

Markers reward the correct general-term and sum formulas and accurate substitution.

Practice4 marks
Paper 2 (complex familiar). A geometric series has first term 2424 and the sum of the first three terms is 4242. (a) Determine the common ratio (given 0<r<10 < r < 1). (b) Determine the sum to infinity.
Show worked answer →

(a) S3=24(1+r+r2)=42S_3 = 24(1 + r + r^2) = 42, so 1+r+r2=1.751 + r + r^2 = 1.75, giving r2+r−0.75=0r^2 + r - 0.75 = 0. Then r=−1±1+32=−1±22r = \dfrac{-1 \pm \sqrt{1 + 3}}{2} = \dfrac{-1 \pm 2}{2}, so r=0.5r = 0.5 (taking the root in 0<r<10 < r < 1).

(b) S∞=a1−r=241−0.5=48.S_\infty = \dfrac{a}{1 - r} = \dfrac{24}{1 - 0.5} = 48.

Markers reward the equation for S3S_3, solving the quadratic for rr, choosing the valid root, and the sum to infinity.

Practise this

Sources & how we know this

ExamExplained