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How do we model situations that change by a constant amount each step?

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

How to identify an arithmetic sequence by its common difference, use the recursive and explicit term rules, sum the terms, and apply the model to simple interest and flat-rate depreciation.

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  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

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.

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.

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.

Exam-style practice questions

Practice questions written in the style of SCSA exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.

WACE 20216 marksA car is bought for \32\,000anddepreciatesbyaflat and depreciates by a flat \24002400 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$?
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 20235 marksThe 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.

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