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Arithmetic and geometric sequences (QCE Math Methods Unit 1)

Syllabus dot point

“Define arithmetic and geometric sequences, find the nnth term and the sum of the first nn terms, and apply to real-world contexts”

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Quick answer

Arithmetic sequences add a common difference dd: Tn=a+(n−1)dT_n = a + (n-1)d and Sn=n2[2a+(n−1)d]S_n = \tfrac{n}{2}[2a+(n-1)d]. Geometric sequences multiply by a common ratio rr: Tn=arn−1T_n = ar^{n-1} and Sn=arn−1r−1S_n = a\dfrac{r^n-1}{r-1}, with S∞=a1−rS_\infty = \dfrac{a}{1-r} when ∣r∣<1|r|<1.

Jump to a section
  1. What this dot point is asking
  2. Arithmetic sequences
  3. Geometric sequences
  4. Identifying the sequence type
  5. Applications
  6. Sum to infinity
  7. In one sentence
  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 you to define and analyse arithmetic and geometric sequences, find the nnth term and the sum of the first nn terms, and apply both to real-world contexts such as loans, salaries, depreciation and population change. Sequences appear in Unit 1 problem-solving and modelling tasks and recur in the financial-mathematics content, so the formulas and the decision of which type applies are foundational skills.

Arithmetic sequences

An arithmetic sequence has a constant common difference dd between consecutive terms, so each term is the previous one plus dd. With first term T1=aT_1 = a,

Tn=a+(n−1)d.T_n = a + (n - 1)d.

The sum of the first nn terms is

Sn=n2[2a+(n−1)d]=n2(T1+Tn),S_n = \frac{n}{2}\big[2a + (n - 1)d\big] = \frac{n}{2}(T_1 + T_n),

the second form being the average of the first and last term multiplied by how many terms there are. The common difference can be found from any two terms by d=Tm−Tkm−kd = \dfrac{T_m - T_k}{m - k}.

Geometric sequences

A geometric sequence has a constant common ratio rr between consecutive terms, so each term is the previous one multiplied by rr. With first term T1=aT_1 = a,

Tn=a rn−1.T_n = a\,r^{n - 1}.

The sum of the first nn terms is

Sn=a rn−1r−1,r≠1.S_n = a\,\frac{r^n - 1}{r - 1}, \qquad r \neq 1.

When ∣r∣<1|r| < 1 the terms shrink toward zero and the infinite series converges:

S∞=a1−r.S_\infty = \frac{a}{1 - r}.

The common ratio is found by dividing any term by the one before it, r=Tn+1Tnr = \dfrac{T_{n+1}}{T_n}.

Identifying the sequence type

The first decision in any sequence problem is which type you have. Compute the differences and ratios of consecutive terms: constant differences signal arithmetic, constant ratios signal geometric, and if neither is constant the sequence is some other type (for example quadratic or recursively defined). Making this check explicitly avoids applying the wrong formula, which is the single most common error.

Applications

The two sequence types model different real-world growth. A fixed-dollar pay rise, straight-line depreciation, and a constant absolute population increase are all arithmetic, because the same amount is added each period. A percentage pay rise, compound interest, declining-balance depreciation, and constant-percentage population growth are all geometric, because the quantity is multiplied by the same factor each period. For compound interest at periodic rate ii, the balance forms a geometric sequence with r=1+ir = 1 + i.

Sum to infinity

For a convergent geometric series (∣r∣<1|r| < 1), the partial sums approach the limit S∞=a1−rS_\infty = \dfrac{a}{1 - r}. This models a total made up of ever-smaller contributions. A ball that rebounds to 80%80\% of its previous height (r=0.8r = 0.8) travels a total downward distance of h1−0.8=5h\dfrac{h}{1 - 0.8} = 5h before coming to rest. The formula fails when ∣r∣≥1|r| \geq 1 because the terms do not shrink and the series diverges.

Sum of a geometric series in an application

Set up the sequence

A patient takes a 200200 mg dose of a drug each day, and each day the body eliminates 25%25\% of whatever is present, leaving 75%75\%. The long-term amount just after a dose is the sum of all remaining contributions, a geometric series with a=200a = 200 and r=0.75r = 0.75.

Apply the sum-to-infinity formula

Since ∣r∣=0.75<1|r| = 0.75 < 1, the series converges:

S∞=a1−r=2001−0.75=2000.25=800 mg.S_\infty = \frac{a}{1 - r} = \frac{200}{1 - 0.75} = \frac{200}{0.25} = 800 \text{ mg}.

Interpret the result

The drug level stabilises at about 800800 mg just after each dose. Markers reward recognising the geometric structure, checking ∣r∣<1|r| < 1 before using S∞S_\infty, and the correct steady-state value with units.

Sum of an arithmetic sequence

Find the sum of the first 2020 terms of 5,8,11,14,…5, 8, 11, 14, \ldots

Step 1: Identify the sequence type and its parameters

Check whether consecutive terms differ by a constant amount. Here 8−5=38 - 5 = 3, 11−8=311 - 8 = 3, 14−11=314 - 11 = 3, so the sequence is arithmetic with first term a=5a = 5 and common difference d=3d = 3.

Step 2: Choose the correct sum formula

Because we know aa, dd, and nn, the most direct formula is:

Sn=n2[2a+(n−1)d]S_n = \frac{n}{2}\big[2a + (n - 1)d\big]

Substitute n=20n = 20, a=5a = 5, d=3d = 3:

S20=202[2(5)+19(3)]S_{20} = \frac{20}{2}\big[2(5) + 19(3)\big]

Step 3: Evaluate inside the bracket first, then multiply

Work from the inside out to avoid errors:

2(5)=10,19(3)=57,10+57=672(5) = 10, \qquad 19(3) = 57, \qquad 10 + 57 = 67

S20=10×67=670S_{20} = 10 \times 67 = 670

Final answer: The sum of the first 2020 terms is 670670.

Common traps
Confusing the two types
Decide whether terms are added or multiplied by checking two consecutive differences and two consecutive ratios before choosing a formula.
Off-by-one on the exponent
The nnth term uses n−1n - 1 (Tn=arn−1T_n = ar^{n-1}), while the sum uses nn (rnr^n in the numerator). Mixing these is a frequent slip.
Using S∞S_\infty when ∣r∣≥1|r| \geq 1
The infinite-sum formula only applies to a convergent series; for ∣r∣≥1|r| \geq 1 the series diverges and the formula is meaningless.
Asking for a term but computing a sum
"The salary in year 1010" is T10T_{10}, a single term; "the total earned over 1010 years" is S10S_{10}. Read which one the question wants.

In one sentence

Arithmetic sequences have common difference dd with Tn=a+(n−1)dT_n = a + (n - 1)d and Sn=n2(T1+Tn)S_n = \tfrac{n}{2}(T_1 + T_n), while geometric sequences have common ratio rr with Tn=arn−1T_n = ar^{n-1}, Sn=arn−1r−1S_n = a\dfrac{r^n - 1}{r - 1}, and S∞=a1−rS_\infty = \dfrac{a}{1 - r} for ∣r∣<1|r| < 1.

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 2 (complex familiar). A worker is paid $45 000\$45\,000 in year 11 with a 4%4\% annual raise. (a) Determine the salary in year 1010. (b) Determine the total earned over the first 1010 years.
Show worked answer →

Geometric sequence with a=45 000a = 45\,000, r=1.04r = 1.04.

(a) T10=45 000×(1.04)9=45 000×1.4233≈$64 049.T_{10} = 45\,000 \times (1.04)^9 = 45\,000 \times 1.4233 \approx \$64\,049.

(b) S10=a r10−1r−1=45 000×1.0410−10.04=45 000×12.0061≈$540 275.S_{10} = a\,\dfrac{r^{10} - 1}{r - 1} = 45\,000 \times \dfrac{1.04^{10} - 1}{0.04} = 45\,000 \times 12.0061 \approx \$540\,275.

Markers reward identifying aa and rr, the correct formula choice, and answers in dollars.

Practice3 marks
Paper 1 (technique). The third term of an arithmetic sequence is 1717 and the seventh term is 3333. (a) Determine the first term and common difference. (b) Determine the sum of the first 2020 terms.
Show worked answer →

(a) T3=a+2d=17T_3 = a + 2d = 17 and T7=a+6d=33T_7 = a + 6d = 33. Subtracting, 4d=164d = 16, so d=4d = 4 and a=17−2(4)=9.a = 17 - 2(4) = 9.

(b) S20=202[2(9)+19(4)]=10[18+76]=10×94=940.S_{20} = \dfrac{20}{2}[2(9) + 19(4)] = 10[18 + 76] = 10 \times 94 = 940.

Markers reward setting up two equations, solving for aa and dd, and the sum-formula evaluation.

Practise this

Sources & how we know this

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