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QLDMath MethodsSyllabus dot point

How are arithmetic and geometric sequences analysed?

Define arithmetic and geometric sequences, find the $n$th term and the sum of the first $n$ terms, and apply to real-world contexts

A focused answer to the QCE Math Methods Unit 1 dot point on sequences. States the $n$th-term and sum formulas for arithmetic ($T_n = a + (n-1)d$) and geometric ($T_n = a r^{n-1}$) sequences, and works the standard QCAA application to debt repayment and salary growth.

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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 to real-world contexts (loans, salaries, depreciation).

Arithmetic sequences

A sequence with a constant common difference dd between consecutive terms.

T1=aT_1 = a, Tn=a+(nβˆ’1)dT_n = a + (n - 1)d.

Sum of first nn terms: Sn=n2[2a+(nβˆ’1)d]=n2(T1+Tn)S_n = \dfrac{n}{2}[2a + (n - 1)d] = \dfrac{n}{2}(T_1 + T_n).

Geometric sequences

A sequence with a constant common ratio rr between consecutive terms.

T1=aT_1 = a, Tn=arnβˆ’1T_n = a r^{n - 1}.

Sum of first nn terms: Sn=arnβˆ’1rβˆ’1S_n = a \dfrac{r^n - 1}{r - 1} for rβ‰ 1r \neq 1.

Sum to infinity (for ∣r∣<1|r| < 1): S∞=a1βˆ’rS_\infty = \dfrac{a}{1 - r}.

Identifying sequence type

Compute consecutive differences and ratios:

  • Constant differences: arithmetic.
  • Constant ratios: geometric.
  • Neither: other type (quadratic, recursive, etc.).

Applications

Loans and savings. Compound interest gives a geometric sequence with r=1+ir = 1 + i where ii is the periodic interest rate.

Depreciation. Straight-line depreciation is arithmetic; declining-balance depreciation is geometric.

Salary growth. Fixed-dollar raise: arithmetic. Percentage raise: geometric.

Population. Constant absolute increase: arithmetic. Constant percentage growth: geometric.

Sum to infinity (geometric series)

For ∣r∣<1|r| < 1, the infinite series converges:

S∞=a+ar+ar2+β‹―=a1βˆ’rS_\infty = a + ar + ar^2 + \cdots = \dfrac{a}{1 - r}.

A bouncing ball that returns 8080% of its previous height (so r=0.8r = 0.8) covers a total of h1βˆ’0.8=5h\dfrac{h}{1 - 0.8} = 5h vertically before stopping.

Worked example

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

Arithmetic with a=5a = 5, d=3d = 3.

S20=202[2(5)+19(3)]=10β‹…[10+57]=10β‹…67=670S_{20} = \dfrac{20}{2}[2(5) + 19(3)] = 10 \cdot [10 + 57] = 10 \cdot 67 = 670.

Common traps

Confusing arithmetic and geometric. Add or multiply? Check by computing two consecutive differences and ratios.

Off-by-one on nβˆ’1n - 1 vs nn. TnT_n uses nβˆ’1n - 1; SnS_n uses nn.

Using sum-to-infinity when ∣r∣β‰₯1|r| \ge 1. Diverges; formula does not apply.

Mixing total and individual term. A question asking for the year 1010 salary wants T10T_{10}, not S10S_{10}.

In one sentence

Arithmetic sequences have common difference dd with Tn=a+(nβˆ’1)dT_n = a + (n - 1)d and Sn=n2(T1+Tn)S_n = \frac{n}{2}(T_1 + T_n); geometric sequences have common ratio rr with Tn=arnβˆ’1T_n = a r^{n-1}, Sn=a(rnβˆ’1)/(rβˆ’1)S_n = a (r^n - 1)/(r - 1), and S∞=a/(1βˆ’r)S_\infty = a/(1 - r) for ∣r∣<1|r| < 1.

Past exam questions, worked

Real questions from past QCAA papers on this dot point, with our answer explainer.

Year 11 SAC4 marksA worker is paid $\$45\,000$ in year $1$ with a $4$% annual raise. Find (a) the salary in year $10$ and (b) the total earned over $10$ years.
Show worked answer β†’

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

(a) Year 1010 salary. T_{10} = 45\,000 \cdot (1.04)^9 = 45\,000 \cdot 1.4233 = \64,049$.

(b) Total over 1010 years. S_{10} = a \dfrac{r^{10} - 1}{r - 1} = 45\,000 \cdot \dfrac{1.04^{10} - 1}{0.04} = 45\,000 \cdot 12.0061 = \540,275$.

Markers reward identification of aa and rr, correct formula choice, and units in dollars.

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