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Arithmetic sequences in WACE Mathematics Applications Unit 3

Syllabus dot point

“Recognise arithmetic sequences, use the recursive and explicit rules, and apply them to simple interest and linear depreciation.”

WACEMathematics ApplicationsUnit 37 min read

Quick answer

An arithmetic sequence changes by a fixed common difference dd each step. Its recursive rule adds dd to the previous term and its explicit rule is tn=a+(n−1)dt_n = a + (n-1)d. It models any quantity that grows or shrinks by a constant amount, such as simple interest and flat-rate depreciation, and its terms sum with Sn=n2(2a+(n−1)d)S_n = \dfrac{n}{2}(2a + (n-1)d).

Jump to a section
  1. What this dot point is asking
  2. The common difference
  3. Where arithmetic sequences appear
  4. Choosing recursive or explicit
  5. Reading an arithmetic model from a worded context
  6. Exam-style questions

What this dot point is asking

You must recognise arithmetic behaviour, use both the recursive and explicit forms, find any term, sum a run of terms, and apply the model to simple interest and straight-line depreciation.

The common difference

A sequence is arithmetic when each term differs from the one before by the same constant, the common difference dd. You test for it by subtracting consecutive terms: if t2−t1=t3−t2=…t_2 - t_1 = t_3 - t_2 = \dots, the sequence is arithmetic. A positive dd gives steady growth (a straight line rising), a negative dd steady decline.

Arithmetic sequence rules

Recursive: tn+1=tn+dt_{n+1} = t_n + d, with first term t1=at_1 = a.

Explicit: tn=a+(n−1)dt_n = a + (n-1)d.

Sum of the first nn terms: Sn=n2(2a+(n−1)d)=n2(a+tn)S_n = \dfrac{n}{2}\big(2a + (n-1)d\big) = \dfrac{n}{2}(a + t_n).

The explicit rule jumps straight to any term without listing the ones before it, which is why it is preferred for the later terms. The sum formula adds a whole run of terms without listing them, useful for total interest or total repayment questions.

Where arithmetic sequences appear

The arithmetic model fits any quantity that changes by a fixed amount per period.

  • Simple interest. Interest is a fixed amount each period because it is always calculated on the original principal, so the balance grows arithmetically.
  • Flat-rate (straight-line) depreciation. An asset loses the same dollar amount each year, so its value falls arithmetically with a negative dd.
  • Regular savings with no interest. Adding a fixed deposit each period builds an arithmetic balance.
Arithmetic means a straight line

Plotting an arithmetic sequence against the step number gives points on a straight line of slope dd. This is the visual signature distinguishing it from the curve of a geometric sequence. The same straight-line behaviour is why simple interest and flat-rate depreciation are called linear models.

Worked example

Simple interest as an arithmetic sequence

$5000\$5000 is invested at 6%6\% simple interest per year. Find the balance after 88 years and the total interest earned.

Step 1: Find the common difference

Simple interest is always calculated on the original principal, so the interest earned each year is the same fixed amount. The interest each year is 6%6\% of $5000=$300\$5000 = \$300, a fixed amount, so the balance is arithmetic with a=5000a = 5000 and d=300d = 300.

Step 2: Write the recursive and explicit rules

Having a constant difference dd means we can describe the balance with both a recursive rule and a direct formula. Recursive rule: Vn+1=Vn+300V_{n+1} = V_n + 300, with V0=5000V_0 = 5000. Explicit rule for the balance after nn years: Vn=5000+300nV_n = 5000 + 300n.

Step 3: Find the balance after 8 years

Substituting n=8n = 8 into the explicit rule gives the balance without needing to list every intermediate year. V8=5000+300×8=5000+2400=7400V_8 = 5000 + 300 \times 8 = 5000 + 2400 = 7400, that is $7400\$7400.

Step 4: Find the total interest

The total interest is the amount gained above the original principal. Total interest is the balance minus the principal: 7400−5000=24007400 - 5000 = 2400, so $2400\$2400.

Final answer: Balance after 88 years is $7400\$7400; total interest earned is $2400\$2400.

Choosing recursive or explicit

Use the recursive rule when generating a table term by term or when technology iterates for you. Use the explicit rule when asked for a single far-off term, such as the value after 3030 years, because it avoids listing every step. Use the sum formula for totals, such as how much is repaid over the life of a flat fortnightly plan.

When you are given two terms (for example t3t_3 and t8t_8) and asked for aa and dd, write both as tn=a+(n−1)dt_n = a + (n-1)d, subtract to eliminate aa and solve for dd, then back-substitute for aa. This two-equation method appears almost every year.

Reading an arithmetic model from a worded context

SCSA questions rarely use the word "arithmetic". You decide the model from the wording. Phrases such as "increases by $40\$40 each month", "loses $2400\$2400 a year" or "adds 250250 litres per day" all signal a fixed step, so the quantity is arithmetic and you set dd to that step (negative if it falls). Phrases such as "increases by 5%5\% each year" signal a fixed multiplier, so the quantity is geometric instead, and you must switch models. Identifying the step versus the multiplier is the single decision that determines every later line of working, so do it before reaching for a formula.

Once the model is fixed, the three tools follow directly: the recursive rule for a table, the explicit rule tn=a+(n−1)dt_n = a + (n-1)d for a distant term, and the sum Sn=n2(2a+(n−1)d)S_n = \dfrac{n}{2}(2a + (n-1)d) for a running total such as total interest, total saved or total distance.

Common errors
Using nn instead of n−1n-1
The explicit rule is tn=a+(n−1)dt_n = a + (n-1)d. Using nn in place of n−1n-1 shifts every term by one dd.
Confusing simple with compound interest
Simple interest adds a fixed amount (arithmetic); compound interest multiplies (geometric). Read which one the question states.
Wrong sign for depreciation
Flat-rate depreciation has a negative common difference because the value falls each year.
Miscounting the first term
Decide whether t1t_1 is the starting value or the value after one period, and keep the count consistent throughout.

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 car is bought for $32 000\$32\,000 and depreciates by a flat $2400\$2400 each year. (a) Write a recursive rule for the value VnV_n after nn years. (b) Find the value after 77 years. (c) After how many complete years does the value first fall below $10 000\$10\,000?
Show worked answer →

Flat-rate depreciation loses the same dollar amount each year, so the value is arithmetic with first term V0=32000V_0 = 32000 and common difference d=−2400d = -2400.

(a) Recursive rule: Vn+1=Vn−2400V_{n+1} = V_n - 2400, with V0=32000V_0 = 32000. (1 mark)

(b) Use the explicit form Vn=32000−2400nV_n = 32000 - 2400n. After 77 years: V7=32000−2400×7=32000−16800=15200V_7 = 32000 - 2400 \times 7 = 32000 - 16800 = 15200, so $15 200\$15\,200. (2 marks)

(c) Solve 32000−2400n<1000032000 - 2400n < 10000, giving 2400n>220002400n > 22000, so n>9.17n > 9.17. The first whole year below $10 000\$10\,000 is n=10n = 10. (3 marks)

Markers reward the negative common difference, a correct explicit substitution, and rounding up to the next whole year for the threshold.

WACE 2023-style5 marks
The third term of an arithmetic sequence is 1717 and the eighth term is 4242. Find the first term and the common difference, then find the sum of the first 2020 terms.
Show worked answer →

Set up two equations from tn=a+(n−1)dt_n = a + (n-1)d.

t3=a+2d=17t_3 = a + 2d = 17 and t8=a+7d=42t_8 = a + 7d = 42. Subtracting: 5d=255d = 25, so d=5d = 5. (2 marks)

Substitute back: a+2(5)=17a + 2(5) = 17, so a=7a = 7. (1 mark)

Sum: S20=202(2(7)+(20−1)(5))=10(14+95)=10×109=1090S_{20} = \dfrac{20}{2}\big(2(7) + (20-1)(5)\big) = 10(14 + 95) = 10 \times 109 = 1090. (2 marks)

Markers reward eliminating aa to find dd first, then a correct sum-formula substitution.

Practise this

Sources & how we know this

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