TASMathematics ApplicationsSyllabus dot point

How do we describe and sum sequences that grow by a constant amount or a constant factor?

Use arithmetic and geometric sequences, including their nth-term and sum rules, to model practical situations.

The nth-term and sum formulas for arithmetic and geometric sequences, with applications to growth, decay, and accumulation in TCE Mathematics Applications.

Generated by Claude Opus 4.78 min answerUpdated 2026-05-30

Reviewed by: AI editorial process; not yet individually human-reviewed

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What this dot point is asking

A sequence is an ordered list of numbers. The growth-and-decay topic uses two named families. This page treats them as sequences in their own right, giving you the explicit term and sum formulas the exam expects you to apply directly.

The nth term

For an arithmetic sequence with first term $a$ and common difference $d$ :

t_n = a + (n-1)d

For a geometric sequence with first term $a$ and common ratio $r$ :

t_n = a r^{\,n-1}

The sum of terms

The sum of the first $n$ terms is written $S_n$ .

Arithmetic series:

S_n = \frac{n}{2}\big(2a + (n-1)d\big) = \frac{n}{2}(a + t_n)

Geometric series (for $r \neq 1$ ):

S_n = \frac{a(r^n - 1)}{r - 1}

Applications

Arithmetic sequences model situations that add a fixed amount each step: simple interest, flat-rate depreciation, or a taxi fare with a flag fall plus a per-kilometre charge. Geometric sequences model situations that multiply by a fixed factor each step: compound interest, population growth at a fixed percentage, or the rebound height of a bouncing ball.

A complete answer states which sequence type applies and why, identifies $a$ together with $d$ or $r$ , substitutes into the correct nth-term or sum formula, and rounds to suit the context.

Exam-style practice questions

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

2024 TASC General Mathematics5 marksA conservationist has bought 200 hectares of land covered in gorse, expecting to clear 12 hectares per year. a) Given the first term is 200, find the next three terms in the sequence representing the uncleared land at the beginning of each year. b) Is this sequence arithmetic or geometric? c) Find the explicit rule for the sequence. d) Use the rule to find the 15th term.

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a) (1 mark) Subtract 12 each year: 188, 176, 164.

b) (1 mark) Arithmetic, because a constant amount (12) is subtracted each year (common difference d = -12).

c) (2 marks) Use Tn = a + (n - 1)d with a = 200 and d = -12: Tn = 200 + (n - 1)(-12), which simplifies to Tn = 212 - 12n.

d) (1 mark) T15 = 212 - 12(15) = 212 - 180 = 32 hectares of uncleared land at the start of the 15th year.

2024 TASC General Mathematics3 marksTasmania's population was 515000 in 2015 (T1) and 572000 in 2022 (T8). Assuming the population follows an arithmetic sequence year by year, find the explicit rule for the sequence. Give values in your final answer to one decimal place.

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(3 marks) The terms run from T1 (2015) to T8 (2022), so there are 7 steps between them.

Common difference d = (T8 - T1) / 7 = (572000 - 515000) / 7 = 57000 / 7 = 8142.9 (1 dp).

Use Tn = a + (n - 1)d with a = 515000: Tn = 515000 + (n - 1)(8142.9). Markers want the common difference computed by dividing by 7 (not 8) and the answer left in explicit form to one decimal place.

2024 TASC General Mathematics7 marksThe explicit rule for an arithmetic sequence is Tn = 22 - 6n. a) The number -260 is a term of this sequence. Determine which term this is. b) If the first term is 16, find the sum of the first 47 terms. c) How many terms of this sequence must be added to achieve a total of -110?

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a) (3 marks) Set Tn = -260: 22 - 6n = -260, so -6n = -282, giving n = 47. It is the 47th term.

b) (2 marks) Here a = T1 = 16 and d = -6. Sum Sn = (n/2)(2a + (n - 1)d) = (47/2)(2 x 16 + 46 x -6) = 23.5 x (32 - 276) = 23.5 x -244 = -5734.

c) (2 marks) Solve Sn = -110: (n/2)(2 x 16 + (n - 1)(-6)) = -110, that is (n/2)(38 - 6n) = -110, giving 19n - 3n squared = -110, so 3n squared - 19n - 110 = 0. Factorising (or the quadratic formula) gives n = 10 (rejecting the negative root). So 10 terms are needed.