Arithmetic and Geometric Sequences - TCE Mathematics Applications (Tasmania)

Syllabus dot point

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

TCEMathematics ApplicationsUnit 38 min readUpdated 2 Jun 2026

Quick answer

You must recognise arithmetic sequences (constant common difference) and geometric sequences (constant common ratio), use their nth-term rules to find any term, and use their sum formulas to total a fixed number of terms.

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.

Arithmetic versus geometric

Arithmetic sequence: each term is the previous term plus a fixed common difference $d$ . Geometric sequence: each term is the previous term times a fixed common ratio $r$ .

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}

Spot the type first

Subtract consecutive terms: a constant gap means arithmetic, and that gap is $d$ . Divide consecutive terms: a constant ratio means geometric, and that ratio is $r$ . Test before choosing a formula.

Finding a term of each type

Two sequences are given and we need to find a specific term in each by identifying the sequence type and applying the correct nth-term formula.

Step 1: Find the 20th term of the arithmetic sequence 5, 8, 11, ...

The sequence increases by 3 each time, so it is arithmetic with first term $a = 5$ and common difference $d = 3$ . Substituting into $t_n = a + (n-1)d$ with $n = 20$ :

t_{20} = 5 + (20-1)(3) = 5 + 57 = 62

Step 2: Find the 8th term of the geometric sequence 3, 6, 12, ...

Each term doubles the one before, so this is geometric with $a = 3$ and common ratio $r = 2$ . Substituting into $t_n = ar^{n-1}$ with $n = 8$ :

t_8 = 3 \times 2^{7} = 3 \times 128 = 384

Final answers: The 20th arithmetic term is 62; the 8th geometric term is 384.

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}

Two forms of the arithmetic sum

Use $S_n = \tfrac{n}{2}(a + t_n)$ when you already know the last term $t_n$ . Use $S_n = \tfrac{n}{2}(2a + (n-1)d)$ when you only know $a$ and $d$ . Both give the same total.

Summing each type

Using the same two sequences from before, we now find the total of a fixed number of terms by applying the appropriate sum formula.

Step 1: Total the first 20 terms of the arithmetic sequence 5, 8, 11, ...

The arithmetic sum formula uses the first term $a = 5$ , the common difference $d = 3$ , and the count $n = 20$ . We use the form that involves $a$ and $d$ because we do not yet know the last term:

S_{20} = \frac{20}{2}\bigl(2 \times 5 + 19 \times 3\bigr) = 10(10 + 57) = 10 \times 67 = 670

Step 2: Total the first 8 terms of the geometric sequence 3, 6, 12, ...

For the geometric series with $a = 3$ , $r = 2$ , and $n = 8$ , we use the geometric sum formula. Because $r \neq 1$ , we divide by $r - 1 = 1$ :

S_8 = \frac{3(2^8 - 1)}{2 - 1} = \frac{3(256 - 1)}{1} = 3 \times 255 = 765

Final answers: The sum of the first 20 arithmetic terms is 670; the sum of the first 8 geometric terms is 765.

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.

Bouncing ball as a geometric sequence

A ball dropped from 2 m rebounds to 70 percent of its previous height each bounce. How high is the fourth rebound?

Identify and apply

This is geometric with $a = 2 \times 0.70 = 1.4$ m for the first rebound and $r = 0.70$ .
Fourth rebound: $t_4 = 1.4 \times 0.70^{\,3} = 1.4 \times 0.343 = 0.4802$ m.

The fourth rebound reaches about 0.48 m. The heights shrink by a constant factor, never quite reaching zero.

Mind the index in the power

In $t_n = a r^{\,n-1}$ the exponent is $n - 1$ , not $n$ , because the first term uses $r^0 = 1$ . A common slip is to raise $r$ to the full $n$ , which gives the next term instead of the one you want.

Using the arithmetic rule on geometric data

If you see the numbers growing fast, check the ratio, not just the difference. Applying $t_n = a + (n-1)d$ to a doubling sequence gives a hopeless underestimate. Always confirm constant difference (arithmetic) or constant ratio (geometric) before picking the formula.

Finding an unknown term number

A common exam task reverses the question: instead of finding a term from its position, you are given a term value and must find which position it is. For an arithmetic sequence, set $t_n = a + (n-1)d$ equal to the known value and solve the linear equation for $n$ . For a geometric sequence you solve $a r^{\,n-1} = \text{value}$ , which usually needs logarithms or a guess-and-check on the calculator. Always check that $n$ comes out as a whole number, because a non-integer means the value is not actually a term of the sequence.

Which term equals a given value

Given the arithmetic sequence with explicit rule $t_n = 212 - 12n$ , we need to find the position of the value $-100$ in the sequence.

Step 1: Set up the equation

To find which term equals $-100$ , replace $t_n$ with $-100$ and solve for $n$ . This works because the explicit rule gives us the value for any position $n$ , so working backwards finds the position for any value:

212 - 12n = -100

Step 2: Solve for n

Rearrange by isolating $n$ on one side:

-12n = -100 - 212 = -312

n = \frac{-312}{-12} = 26

Step 3: Check the result is valid

Since $n = 26$ is a positive whole number, the value $-100$ does appear in the sequence. It is the 26th term. If the algebra had produced a fraction or a negative, $-100$ would not be a term in this sequence at all.

Final answer: $-100$ is the 26th term.

Infinite geometric series

When the common ratio satisfies $|r| < 1$ , the terms shrink toward zero and the running total approaches a finite limit even if you add forever. This limiting sum is

S_\infty = \frac{a}{1 - r}, \quad \text{valid only when } |r| < 1.

This appears in TASC questions asking what a continually decaying quantity totals "in the long run". If

|r| \ge 1

, the terms do not shrink and the infinite sum has no finite value.

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, checks that any term number $n$ is a whole number, and rounds to suit the context.

Exam-style questions

Questions in the style of TASC 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.

Practice5 marks

A conservationist buys 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 for the uncleared land at the start of each year. b) Is this sequence arithmetic or geometric? c) Find the explicit rule. d) Use the rule to find the 15th term.

Show worked answer →

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

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

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

d) (1 mark) $t_{15} = 212 - 12(15) = 212 - 180 = 32$ hectares uncleared at the start of the 15th year.

Practice3 marks

Tasmania's population was

515\,000

in 2015 (

t_1

) and

572\,000

in 2022 (

t_8

). Assuming the population follows an arithmetic sequence year by year, find the explicit rule. Give values to one decimal place.

Show worked answer →

(3 marks) The terms run from $t_1$ (2015) to $t_8$ (2022), so there are 7 steps between them.

Common difference $d = \frac{t_8 - t_1}{7} = \frac{572000 - 515000}{7} = \frac{57000}{7} = 8142.9$ (1 dp).

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

Practice7 marks

The explicit rule for an arithmetic sequence is

t_n = 22 - 6n

. a) The number

-260

is a term of this sequence. Which term is it? b) If the first term is 16, find the sum of the first 47 terms. c) How many terms must be added to achieve a total of

-110

Show worked answer →

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

b) (2 marks) Here $a = t_1 = 16$ and $d = -6$ . Sum $S_n = \frac{n}{2}(2a + (n - 1)d) = \frac{47}{2}(2 \times 16 + 46 \times (-6)) = 23.5 \times (32 - 276) = 23.5 \times (-244) = -5734$ .

c) (2 marks) Solve $S_n = -110$ : $\frac{n}{2}(2 \times 16 + (n - 1)(-6)) = -110$ , that is $\frac{n}{2}(38 - 6n) = -110$ , giving $19n - 3n^2 = -110$ , so $3n^2 - 19n - 110 = 0$ . Factorising (or the quadratic formula) gives $n = 10$ (rejecting the negative root). So 10 terms are needed.

Practise this

Sources & how we know this

General Mathematics Level 3 (MTG315123) course document (2023 onwards) — TASC