Arithmetic sequences and series for HSC Maths Advanced financial modelling

Syllabus dot point

“Use the formulas for the nth term and the sum of n terms of an arithmetic sequence in financial and modelling contexts”

HSCMaths AdvancedYear 12: Financial Mathematics14 min readUpdated 20 Jun 2026

Quick answer

An arithmetic sequence has $T_n = a + (n - 1)d$ with partial sum $S_n = \frac{n}{2}(2a + (n - 1)d) = \frac{n}{2}(a + l)$ , and this constant-difference model underlies simple interest, salary increments, straight-line depreciation and stepped repayment schedules.

Jump to a section

What this dot point is asking
The answer
Practice questions

What this dot point is asking

NESA wants you to recognise arithmetic sequences and series, apply the formula for the $n$ th term $T_n = a + (n - 1)d$ and the two partial-sum formulas, and use them in financial and modelling contexts such as simple interest, salary increments, straight-line depreciation and stacked schedules.

Note

Think of climbing a staircase where every step is exactly the same height. To know how high you are after some steps, you multiply the step height by how many steps you have climbed and add your starting height. That constant step is what makes a sequence arithmetic: you add the same amount each time, never multiply. Money behaves this way whenever the change is a fixed dollar figure rather than a percentage: simple interest that pays the same dollars every year, a pay rise of the same amount each year, or a machine that loses the same value annually. Adding up all those evenly-spaced amounts is an arithmetic series, and there is a neat shortcut: pair the first amount with the last, and you get the same total every time.

The answer

An arithmetic sequence is what you get whenever a quantity changes by the same fixed amount each period. The moment the change is a constant number of dollars, degrees or units, a fixed simple-interest payment, an equal annual pay rise, a fixed depreciation charge, you have an arithmetic sequence, and its running total is an arithmetic series. This is the constant-difference cousin of the geometric (constant-ratio) sequences that drive compound interest. The whole topic reduces to two ideas: the $n$ th term $T_n = a + (n - 1)d$ , and the sum of the first $n$ terms, which has two equivalent forms.

Arithmetic sequences

An arithmetic sequence has a constant common difference $d$ between consecutive terms. With first term $a$ ,

T_n = a + (n - 1)d.

So $T_1 = a$ , $T_2 = a + d$ , $T_3 = a + 2d$ , and so on: you add $d$ exactly $n - 1$ times to reach the $n$ th term. The defining test is that subtracting any term from the next always gives the same number $d$ (compare a geometric sequence, where you divide to get a constant ratio). Because each term rises by the same step, the terms plotted against their position sit on a straight line of gradient $d$ .

Arithmetic series (partial sum)

The sum of the first $n$ terms is

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

There is an equivalent form that uses the last term $l = T_n$ instead of $d$ :

S_n = \frac{n}{2}(a + l).

The two are the same formula: substituting $l = a + (n - 1)d$ into the second gives $\frac{n}{2}(a + a + (n - 1)d) = \frac{n}{2}(2a + (n - 1)d)$ . Use the first when you know $a$ and $d$ ; use the $\frac{n}{2}(a + l)$ form when you already know (or can quickly find) both ends, because averaging the two ends and multiplying by the number of terms is faster and less error-prone.

Why the sum formula works

The $\frac{n}{2}(a + l)$ form is the famous pairing trick. Write the sum forwards and backwards and add:

\begin{aligned} S_n &= a + (a + d) + \cdots + (l - d) + l \\ S_n &= l + (l - d) + \cdots + (a + d) + a. \end{aligned}

Each vertical pair adds to $a + l$ , and there are $n$ pairs, so $2 S_n = n(a + l)$ , giving $S_n = \frac{n}{2}(a + l)$ . The following build-up shows the partial sum growing one term at a time for the sequence $3, 5, 7, 9, 11$ .

Key fact

For an arithmetic sequence with first term $a$ and common difference $d$ : the $n$ th term is $T_n = a + (n - 1)d$ ; the sum of the first $n$ terms is $S_n = \dfrac{n}{2}\big(2a + (n - 1)d\big) = \dfrac{n}{2}(a + l)$ , where $l = T_n$ is the last term. The defining feature is a constant difference (you add $d$ each step), not a constant ratio. Always identify $a$ and $d$ first: subtract consecutive terms to get $d$ , and remember the $n$ th term adds $d$ exactly $n - 1$ times, not $n$ times.

Simple interest as an arithmetic sequence

Simple interest is charged only on the original principal $P$ , so each period adds the same fixed amount $I = P r$ (with $r$ the per-period rate as a decimal). The totals $P + Pr, P + 2Pr, P + 3Pr, \dots$ form an arithmetic sequence with common difference $Pr$ . The amount after $n$ periods is

A_n = P(1 + rn),

a straight line in $n$ . This is exactly what separates simple interest (constant difference, arithmetic) from compound interest (constant ratio, geometric).

Straight-line depreciation

An asset that loses a fixed dollar amount $D$ each year (the "prime cost" or straight-line method) has values $V_0, V_0 - D, V_0 - 2D, \dots$ , an arithmetic sequence with common difference $-D$ . The value after $n$ years is $V_n = V_0 - Dn$ . Contrast this with declining-balance depreciation, which loses a fixed percentage each year and is therefore geometric.

Salary increments and equal instalments

A salary that rises by the same dollar amount each year, an allowance topped up by a fixed sum each period, or a repayment plan that steps up (or down) by an equal amount are all arithmetic. To find a total earned or paid over $n$ periods, use the partial sum $S_n$ ; to find the amount in a particular period, use $T_n$ .

How exam questions ask about arithmetic sequences and series

The context is dressed up, but each version reduces to identifying $a$ and $d$ and choosing $T_n$ or $S_n$ :

"Find the value after $n$ years" of an asset losing a fixed dollar amount each year. Straight-line depreciation: $V_n = V_0 - Dn$ , an arithmetic sequence with difference $-D$ .
"Find the $n$ th term" or "which term equals $X$ ?" Use $T_n = a + (n - 1)d$ ; for "which term", set it equal to $X$ and solve the linear equation for $n$ .
"Find the sum of the first $n$ terms" or a total of stepped payments. Partial sum $S_n = \dfrac{n}{2}(2a + (n - 1)d)$ , or $\dfrac{n}{2}(a + l)$ if both ends are known.
"Show that the totals form an arithmetic sequence." Subtract consecutive values to show the difference is constant, then state $a$ and $d$ ; markers want the constant difference demonstrated, not assumed.
"How many terms until the total reaches $X$ ?" Set $S_n = X$ , form a quadratic in $n$ , solve and round to the sensible whole number of terms with a check.

Worked example

Four calculations showing the arithmetic formulas applied to a standard sequence, a simple-interest total, a straight-line depreciation model and a "how many terms" sum.

Compute $T_{12}$ and $S_{12}$ for the sequence $4, 9, 14, 19, \dots$

Each term is $5$ more than the previous, so $a = 4$ and $d = 5$ .

The $n$ th-term formula gives:

T_{12} = a + (n - 1)d = 4 + (12 - 1)(5) = 4 + 55 = 59.

The partial-sum formula gives:

S_{12} = \frac{n}{2}\big(2a + (n - 1)d\big) = \frac{12}{2}\big(2(4) + 11(5)\big) = 6(8 + 55) = 6 \times 63 = 378.

Total a simple-interest investment

$8000 earns simple interest at $5\%$ per annum. The yearly interest is a fixed $8000 \times 0.05 = 400$ , so the totals form an arithmetic sequence with $a = 8400$ (after year $1$ ) and $d = 400$ . The amount after $n$ years is

A_n = P(1 + rn) = 8000(1 + 0.05 n).

After $9$ years,

A_9 = 8000(1 + 0.05 \times 9) = 8000(1 + 0.45) = 8000 \times 1.45 = 11600.

So the investment is worth $11600 after $9$ years.

Apply the straight-line depreciation model

A van bought for $45000 depreciates by a fixed $5000 each year. The values form an arithmetic sequence with $a = 45000$ (at $n = 0$ ) and $d = -5000$ , so

V_n = 45000 - 5000n.

After $7$ years,

V_7 = 45000 - 5000(7) = 45000 - 35000 = 10000.

So the van is worth $10000 after $7$ years.

Find how many terms sum to a target

For the series $6 + 10 + 14 + \cdots$ (so $a = 6$ , $d = 4$ ), find $n$ so that $S_n = 510$ .

S_n = \frac{n}{2}\big(2(6) + (n - 1)(4)\big) = \frac{n}{2}(4n + 8) = 2n^2 + 4n.

2n^2 + 4n = 510 \quad\Rightarrow\quad n^2 + 2n - 255 = 0 \quad\Rightarrow\quad (n + 17)(n - 15) = 0.

Rejecting the negative root, $n = 15$ . Check: $S_{15} = 2(225) + 4(15) = 450 + 60 = 510$ . So $15$ terms sum to $510$ .

Final answers: $T_{12} = 59$ , $S_{12} = 378$ ; simple-interest total $11600; van value $10000; $n = 15$ terms.

Common traps

Off-by-one on the term count: $T_n = a + (n - 1)d$ adds $d$ exactly $n - 1$ times, not $n$ times. The $10$ th term of $3, 7, 11, \dots$ is $3 + 9 \times 4 = 39$ , not $3 + 10 \times 4$ .
Arithmetic mistaken for geometric: A constant difference (you add $d$ ) is arithmetic; a constant ratio (you multiply by $r$ ) is geometric. Simple interest and straight-line depreciation are arithmetic; compound interest and declining-balance depreciation are geometric.
Difference read as a percentage: In straight-line depreciation the asset loses a fixed dollar amount, so $d$ is a subtraction of dollars, not a percentage of the current value. A fixed $1800 per year gives $d = -1800$ , full stop.
Dropping the outer $\frac{n}{2}$ in the sum: $S_n = \frac{n}{2}(2a + (n - 1)d)$ : the whole bracket is halved and multiplied by $n$ . Writing $S_n = 2a + (n - 1)d$ (just the bracket) is a frequent slip.
Fractional number of terms: When solving $S_n = X$ you get a quadratic; $n$ must be a whole number of terms. If the root is not an integer, check the terms either side rather than quoting a fraction.

Exam technique

State $a$ and $d$ as your first line; almost every method mark hangs on identifying them correctly, and $d$ is found by subtracting consecutive terms. When both ends of a series are known, reach for $S_n = \frac{n}{2}(a + l)$ (average the ends, multiply by the count) rather than the longer form. For "which term equals $X$ ", set $T_n = X$ and solve a linear equation; for "how many terms sum to $X$ ", set $S_n = X$ and solve a quadratic, then round to a whole number of terms with a check. Always confirm a sequence is arithmetic by showing the difference is constant before quoting a formula, because "show that" awards a mark for that demonstration.

Practice questions

Original practice questions graded from foundation to exam level, each with a full worked solution. Try them before revealing the solution.

foundation2 marks

For the arithmetic sequence

7, 11, 15, 19, \dots

find

T_{20}

Show worked solution →

Identify the first term and common difference. Subtracting consecutive terms, $11 - 7 = 4$ , so $a = 7$ and $d = 4$ .

Apply the $n$ th-term formula.

T_{20} = a + (n - 1)d = 7 + (20 - 1)(4) = 7 + 19 \times 4 = 7 + 76 = 83.

Marker's note: one mark for identifying $a = 7$ and $d = 4$ , one for $T_{20} = 83$ from the correct formula. Using $(n)d = 20 \times 4$ instead of $(n - 1)d = 19 \times 4$ is the usual off-by-one slip.

foundation2 marks

Find the sum of the first

30

terms of the arithmetic series

2 + 5 + 8 + 11 + \cdots

Show worked solution →

Read off $a$ and $d$ . The first term is $a = 2$ and $d = 5 - 2 = 3$ .

Apply the partial-sum formula.

S_{30} = \frac{n}{2}\big(2a + (n - 1)d\big) = \frac{30}{2}\big(2(2) + (30 - 1)(3)\big) = 15\big(4 + 87\big) = 15 \times 91 = 1365.

Marker's note: one mark for the correct substitution into $S_n = \frac{n}{2}(2a + (n - 1)d)$ , one for $S_{30} = 1365$ . Forgetting the outer factor $\frac{n}{2}$ is the standard error.

foundation3 marks

A machine bought for $24000 is depreciated by the straight-line method at a fixed $1800 per year. (a) Write a formula for its value

V_n

after

n

years. (b) Find its value after

6

years.

Show worked solution →

Part (a): recognise a fixed yearly drop as an arithmetic sequence. The value starts at $24000 and falls by the same $1800 each year, so the common difference is $d = -1800$ . Taking the purchase value as the term at $n = 0$ ,

V_n = 24000 - 1800n.

Part (b): substitute $n = 6$ .

V_6 = 24000 - 1800(6) = 24000 - 10800 = 13200.

So the machine is worth $13200 after $6$ years.

Marker's note: one mark for the difference $d = -1800$ (a subtraction, not a percentage), one for a correct linear formula, one for $V_6 = 13200$ . Multiplying by a rate rather than subtracting a fixed amount confuses straight-line with declining-balance depreciation.

core3 marks

A principal of $5000 earns simple interest at

6\%

per annum. Let

A_n

be the total amount (principal plus interest) after

n

years. (a) Show that the yearly totals form an arithmetic sequence and state

a

and

d

. (b) Find the total after

12

years.

Show worked solution →

Part (a): compute the yearly interest. Simple interest pays a fixed $6\%$ of the original principal each year:

I = 5000 \times 0.06 = 300 \text{ per year}.

The totals are $5300, $5600, $5900, and so on: each year adds the same $300, so the totals form an arithmetic sequence. Taking the total after $1$ year as the first term, $a = 5300$ and $d = 300$ .

Part (b): find the total after $12$ years using $A_n = P(1 + rn)$ . The simple-interest amount is

A_{12} = 5000\big(1 + 0.06 \times 12\big) = 5000(1 + 0.72) = 5000 \times 1.72 = 8600.

So the total after $12$ years is $8600.

Marker's note: one mark for the constant yearly interest $300 (fixed, because simple interest is charged on the original principal only), one for stating $a$ and $d$ , one for $A_{12} = 8600$ . Treating the interest as compounding (a constant ratio rather than a constant difference) is the trap.

core4 marks

A graduate scheme lists a salary schedule in a table. Year 1 pays 62000, Year 2 pays 65200, Year 3 pays 68400 and Year 4 pays 71600 (all in dollars). (a) Show that the salaries form an arithmetic sequence and state the annual rise. (b) Find the salary in Year 10. (c) Find the total earned over the first 10 years.

Show worked solution →

Part (a): test for a constant difference. Subtract consecutive salaries:

65200 - 62000 = 3200, \qquad 68400 - 65200 = 3200, \qquad 71600 - 68400 = 3200.

The difference is constant at $3200, so the salaries form an arithmetic sequence with $a = 62000$ and $d = 3200$ .

Part (b): the Year 10 salary is $T_{10}$ .

T_{10} = a + (n - 1)d = 62000 + (10 - 1)(3200) = 62000 + 9 \times 3200 = 62000 + 28800 = 90800.

Part (c): the total earned is $S_{10}$ . Using the last-term form with $l = T_{10} = 90800$ ,

S_{10} = \frac{n}{2}(a + l) = \frac{10}{2}(62000 + 90800) = 5 \times 152800 = 764000.

So the salary in Year 10 is $90800 and the ten-year total is $764000.

Marker's note: one mark for showing the constant difference $3200, one for $T_{10} = 90800$ , one for the correct sum method, one for the total $764000. Using $9$ terms instead of $10$ , or the rise as a percentage, are the common slips.

core3 marks

The sum of the first

n

terms of an arithmetic series is

S_n = 100

for the series

4 + 7 + 10 + \cdots

. Find

n

Show worked solution →

Set up $S_n$ with $a = 4$ and $d = 3$ .

S_n = \frac{n}{2}\big(2a + (n - 1)d\big) = \frac{n}{2}\big(8 + 3(n - 1)\big) = \frac{n}{2}(3n + 5).

Solve $S_n = 100$ .

\frac{n}{2}(3n + 5) = 100 \quad\Rightarrow\quad n(3n + 5) = 200 \quad\Rightarrow\quad 3n^2 + 5n - 200 = 0.

Use the quadratic formula.

n = \frac{-5 \pm \sqrt{25 + 2400}}{6} = \frac{-5 \pm \sqrt{2425}}{6} = \frac{-5 \pm 49.244\ldots}{6}.

The positive root is $n = \dfrac{44.244\ldots}{6} \approx 7.37$ . Since $n$ must be a whole number of terms and $S_7 = 91$ while $S_8 = 116$ , no whole $n$ gives exactly $100$ ; the first total to reach or exceed $100$ is at $n = 8$ .

Marker's note: one mark for the quadratic $3n^2 + 5n - 200 = 0$ , one for solving it, one for interpreting the non-integer root sensibly (checking $S_7$ and $S_8$ rather than quoting a fractional number of terms). Accepting $n \approx 7.37$ as a final answer ignores that $n$ counts terms.

exam5 marks

Each month Mia adds $50 more to her repayment than the month before, starting with $200 in month 1. (a) Show that her monthly repayments form an arithmetic sequence and write the amount paid in month

n

. (b) Find the total repaid after 24 months. (c) The loan plus interest totals $30000. In which month does the running total first reach or exceed $30000?

Show worked solution →

Part (a): a fixed monthly increase is an arithmetic sequence. The first repayment is $a = 200$ and each month adds a constant $50, so $d = 50$ and

T_n = a + (n - 1)d = 200 + 50(n - 1) = 150 + 50n.

Part (b): the 24-month total is $S_{24}$ .

S_{24} = \frac{n}{2}\big(2a + (n - 1)d\big) = \frac{24}{2}\big(2(200) + 23(50)\big) = 12\big(400 + 1150\big) = 12 \times 1550 = 18600.

So $18600 is repaid after $24$ months.

Part (c): solve $S_n \ge 30000$ . With $S_n = \dfrac{n}{2}(400 + 50(n - 1)) = \dfrac{n}{2}(50n + 350) = 25n^2 + 175n$ ,

25n^2 + 175n \ge 30000 \quad\Rightarrow\quad n^2 + 7n - 1200 \ge 0.

Solving the boundary quadratic,

n = \frac{-7 + \sqrt{49 + 4800}}{2} = \frac{-7 + \sqrt{4849}}{2} = \frac{-7 + 69.635\ldots}{2} \approx 31.3.

The running total is below $30000 at $n = 31$ and above it at $n = 32$ . Check: $S_{31} = 25(961) + 175(31) = 24025 + 5425 = 29450$ , and $S_{32} = 25(1024) + 175(32) = 25600 + 5600 = 31200$ . So the total first reaches $30000 in month 32.

Marker's note: one mark for $d = 50$ and the formula for $T_n$ , one for $S_{24} = 18600$ , one for forming the inequality $n^2 + 7n - 1200 \ge 0$ , one for solving it, one for rounding up with a check to month $32$ . Rounding $31.3$ down to $31$ (where the total is still below target) is the standard error.

exam4 marks

Logs are stacked so that the bottom row holds

30

logs, and each row above holds

2

fewer than the row below. The stack ends with a top row of

6

logs. (a) How many rows are in the stack? (b) How many logs are in the stack altogether?

Show worked solution →

Part (a): the row counts form an arithmetic sequence. From the bottom up, $a = 30$ and each row loses $2$ , so $d = -2$ . The top row has $6$ logs, so it is the term equal to $6$ :

T_n = a + (n - 1)d = 30 + (n - 1)(-2) = 6.

30 - 2(n - 1) = 6 \quad\Rightarrow\quad 2(n - 1) = 24 \quad\Rightarrow\quad n - 1 = 12 \quad\Rightarrow\quad n = 13.

So there are $13$ rows.

Part (b): total logs is $S_{13}$ . Using the first-and-last form with $a = 30$ and $l = 6$ ,

S_{13} = \frac{n}{2}(a + l) = \frac{13}{2}(30 + 6) = \frac{13}{2} \times 36 = 13 \times 18 = 234.

So the stack holds $234$ logs.

Marker's note: one mark for $d = -2$ , one for solving $T_n = 6$ to get $n = 13$ , one for the correct sum formula, one for the total $234$ . The $\frac{n}{2}(a + l)$ form is fastest here because both ends are known; using $\frac{n}{2}(2a + (n - 1)d)$ works too but invites arithmetic slips.

Practise this

Sources & how we know this

Mathematics Advanced Stage 6 Syllabus (2017) — NESA
Mathematics Advanced HSC exam papers — NESA

What this dot point is asking

The answer

Arithmetic sequences

Arithmetic series (partial sum)

Why the sum formula works

Simple interest as an arithmetic sequence

Straight-line depreciation

Salary increments and equal instalments

How exam questions ask about arithmetic sequences and series

Compute T12T_{12}T12​ and S12S_{12}S12​ for the sequence 4,9,14,19,…4, 9, 14, 19, \dots4,9,14,19,…

Total a simple-interest investment

Apply the straight-line depreciation model

Find how many terms sum to a target

Practice questions

Practise this

Sources & how we know this

Compute $T_{12}$ and $S_{12}$ for the sequence $4, 9, 14, 19, \dots$