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Geometric sequences in WACE Mathematics Applications Unit 3

Syllabus dot point

“Recognise geometric sequences, use the recursive and explicit rules with the common ratio, and apply them to growth and decay.”

WACEMathematics ApplicationsUnit 37 min read

Quick answer

A geometric sequence multiplies by a fixed common ratio RR each step, so it models constant-percentage change. The recursive rule is tn+1=R tnt_{n+1} = R\,t_n and the explicit rule is tn=aRn−1t_n = a R^{n-1}. A ratio above 11 gives growth (populations, compound interest); a ratio between 00 and 11 gives decay (reducing-balance depreciation, half-life). Find the ratio by dividing consecutive terms.

Jump to a section
  1. What this dot point is asking
  2. The common ratio
  3. Growth versus decay
  4. Switching between the forms
  5. Doubling and halving times
  6. Comparing arithmetic and geometric growth
  7. Exam-style questions

What this dot point is asking

You must recognise geometric behaviour, find the common ratio, use both forms of the rule, and apply the model to percentage growth and decay.

The common ratio

A sequence is geometric when each term is a fixed multiple of the one before, the common ratio RR. Test by dividing consecutive terms: if t2t1=t3t2=…\dfrac{t_2}{t_1} = \dfrac{t_3}{t_2} = \dots, the sequence is geometric.

Geometric sequence rules

Recursive: tn+1=R tnt_{n+1} = R\,t_n, with first term t1=at_1 = a.

Explicit: tn=aRn−1t_n = a R^{n-1}.

Common ratio from a percentage change: R=1+percentage100R = 1 + \dfrac{\text{percentage}}{100} (growth) or R=1−percentage100R = 1 - \dfrac{\text{percentage}}{100} (decay).

A growth of r%r\% per period gives R=1+r100R = 1 + \dfrac{r}{100}; a decay of r%r\% gives R=1−r100R = 1 - \dfrac{r}{100}. So 8%8\% growth is R=1.08R = 1.08 and 12%12\% decay is R=0.88R = 0.88.

Growth versus decay

The value of RR decides the whole shape of the model.

  • R>1R > 1: geometric growth, an upward curve that steepens (compound interest, populations).
  • 0<R<10 < R < 1: geometric decay, a downward curve that flattens towards zero (reducing-balance depreciation, radioactive decay).
  • R=1R = 1: a constant sequence; nothing changes.
Geometric means a curve, arithmetic means a line

Plotting a geometric sequence gives a curve (exponential), not the straight line of an arithmetic sequence. The signal of a geometric model in a worded question is a percentage change ("rises by 5%5\%"), whereas a fixed dollar change ("rises by $50\$50") signals arithmetic.

Worked example

Reducing-balance depreciation as a geometric sequence

A van bought for $60 000\$60\,000 loses 20%20\% of its value each year. Find its value after 55 years and the year it first falls below $20 000\$20\,000.

Step 1: Find the common ratio

A percentage loss each year produces a constant multiplier, which is the defining feature of a geometric sequence. A 20%20\% annual loss multiplies by 1−0.20=0.801 - 0.20 = 0.80, so R=0.80R = 0.80 and a=60000a = 60000.

Step 2: Write the explicit rule

With the starting value and the common ratio known, the explicit rule gives any future value directly without stepping through every year. Vn=60000×0.80nV_n = 60000 \times 0.80^n, where nn is the number of years.

Step 3: Find the value after 5 years

Substituting n=5n = 5 into the explicit rule gives the value at that point in time. V5=60000×0.805=60000×0.32768=19660.80V_5 = 60000 \times 0.80^5 = 60000 \times 0.32768 = 19660.80, that is $19 660.80\$19\,660.80.

Step 4: Find the year it first falls below $20 000\$20\,000

To find when a threshold is crossed, set up the inequality and test integer values of nn. Solve 60000×0.80n<2000060000 \times 0.80^n < 20000, so 0.80n<0.33330.80^n < 0.3333. Testing, 0.804=0.40960.80^4 = 0.4096 and 0.805=0.32770.80^5 = 0.3277, so the value first falls below $20 000\$20\,000 during year 55.

Final answer: Value after 55 years is $19 660.80\$19\,660.80; the value first falls below $20 000\$20\,000 during year 55.

Switching between the forms

Use the recursive rule to build a table or to let technology iterate; use the explicit rule tn=aRn−1t_n = a R^{n-1} for a single far-off term. When given two terms, divide them to find RR: if t2=aRt_2 = aR and t5=aR4t_5 = aR^4, then t5t2=R3\dfrac{t_5}{t_2} = R^3, so R=t5t23R = \sqrt[3]{\dfrac{t_5}{t_2}}, and back-substitute for aa.

Note the index convention: tn=aRn−1t_n = a R^{n-1} uses n−1n-1 when t1=at_1 = a is the first term. In finance the version An=A0RnA_n = A_0 R^n is common, where A0A_0 is the starting value before any period. Decide which counting you are using and stay consistent.

Doubling and halving times

A common question asks how long a geometric model takes to double (for growth) or halve (for decay). Set the explicit rule equal to twice (or half) the start and solve by trial on the calculator. For 8%8\% annual growth, solve 1.08n=21.08^n = 2: testing gives 1.089=1.9991.08^9 = 1.999, so doubling takes about 99 years. For a decay with R=0.85R = 0.85, solve 0.85n=0.50.85^n = 0.5: 0.854=0.5220.85^4 = 0.522 and 0.855=0.4440.85^5 = 0.444, so the quantity halves during year 55. Reporting the first whole period that crosses the target is the expected form of the answer.

Comparing arithmetic and geometric growth

Over a few periods, a steep arithmetic increase can stay ahead of a modest geometric one, but geometric growth always overtakes eventually because it compounds. This is the reason compound interest beats simple interest over time and why a small percentage population growth produces large absolute increases in later years. When a question gives both an "increases by a fixed amount" option and an "increases by a percentage" option, expect to model one as arithmetic and the other as geometric and compare them at a chosen period.

Common errors
Adding instead of multiplying
Geometric sequences multiply by RR; using a common difference makes it arithmetic and gives a wrong answer.
Wrong ratio from a percentage
A 15%15\% decay is R=0.85R = 0.85, not R=0.15R = 0.15 or R=−0.15R = -0.15.
Index confusion
tn=aRn−1t_n = a R^{n-1} uses n−1n-1 when the first term is t1t_1; check whether the problem counts from 00 or 11.
Forgetting to round only at the end
Keep full precision in RnR^n and round the final answer, especially for dollar values.

Exam-style questions

Questions in the style of SCSA 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.

WACE 2021-style6 marks
A town's population is 24 00024\,000 and grows by 3%3\% each year. (a) Write a recursive rule for the population PnP_n after nn years. (b) Find the population after 1010 years. (c) In how many complete years does the population first exceed 32 00032\,000?
Show worked answer →

Growth of 3%3\% per year multiplies by 1.031.03, so the model is geometric.

(a) Common ratio R=1.03R = 1.03, so Pn+1=1.03 PnP_{n+1} = 1.03\,P_n, with P0=24000P_0 = 24000. (1 mark)

(b) Explicit form Pn=24000×1.03nP_n = 24000 \times 1.03^n. After 1010 years: P10=24000×1.0310=24000×1.34392=32254P_{10} = 24000 \times 1.03^{10} = 24000 \times 1.34392 = 32254, so about 32 25432\,254 people. (2 marks)

(c) Solve 24000×1.03n>3200024000 \times 1.03^n > 32000, so 1.03n>1.33331.03^n > 1.3333. By trial, 1.039=1.30481.03^9 = 1.3048 and 1.0310=1.34391.03^{10} = 1.3439, so the population first exceeds 32 00032\,000 in year 1010. (3 marks)

Markers reward the ratio 1.031.03, a correct explicit substitution, and finding the first whole year above the threshold.

WACE 2023-style5 marks
A machine bought for $45 000\$45\,000 depreciates by 15%15\% of its value each year. (a) State the common ratio. (b) Find its value after 44 years. (c) Find the percentage of the original value remaining after 44 years.
Show worked answer →

A 15%15\% loss each year multiplies by 0.850.85, a geometric decay.

(a) Common ratio R=1−0.15=0.85R = 1 - 0.15 = 0.85. (1 mark)

(b) Vn=45000×0.85nV_n = 45000 \times 0.85^n, so V4=45000×0.854=45000×0.52200=23490V_4 = 45000 \times 0.85^4 = 45000 \times 0.52200 = 23490, that is $23 490\$23\,490. (2 marks)

(c) Fraction remaining =0.854=0.522=52.2%= 0.85^4 = 0.522 = 52.2\% of the original value. (2 marks)

Markers reward the decay ratio 0.850.85, the explicit substitution, and reading the remaining fraction directly from RnR^n.

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