Geometric 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 a geometric sequence, and the limiting sum, in financial contexts”

HSCMaths AdvancedYear 12: Financial Mathematics14 min readUpdated 20 Jun 2026

Quick answer

A geometric sequence has $T_n = a r^{n - 1}$ with finite sum $S_n = \frac{a(1 - r^n)}{1 - r}$ and limiting sum $\frac{a}{1 - r}$ when $|r| < 1$ , and these formulas underlie compound interest, depreciation, annuities and perpetuities.

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What this dot point is asking
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Exam-style questions
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What this dot point is asking

NESA wants you to recognise geometric sequences and series, apply the formulas for the $n$ th term, the sum of $n$ terms and the limiting sum, and use them in financial contexts such as repeated payments, depreciation, and perpetuities.

Note

Think about forwarding a chain message to the same number of friends each round. If everyone sends it to $3$ people, the count of new readers multiplies by $3$ every step: $1$ , then $3$ , then $9$ , then $27$ . That constant multiplier is what makes a sequence geometric, and it is exactly how money behaves when a bank multiplies your balance by the same interest factor every year, or a car loses the same slice of its value each year. Adding up all those rounds gives a geometric series. The surprising part is that when the multiplier is a fraction smaller than $1$ , so each round is smaller than the last, an endless list of payments can still add up to one sensible finite total.

The answer

Geometric sequences are the engine behind every formula in this topic. The moment a quantity is repeatedly multiplied by the same factor each period, a fixed interest rate, a fixed depreciation rate, a stream of equal payments each discounted by one more period, you have a geometric sequence, and its sum is a geometric series. Compound interest, declining-balance depreciation, the future value of an annuity and the present value of a perpetuity are all the same three formulas wearing different clothes. The one genuinely new idea is the limiting sum: when the multiplier has size less than one, an infinite list of terms can still add to a finite total.

Geometric sequences

A geometric sequence has a constant ratio $r$ between consecutive terms. With first term $a$ ,

T_n = a r^{n - 1}.

So $T_1 = a$ , $T_2 = a r$ , $T_3 = a r^2$ , and so on. The defining test is that dividing any term by the one before it always gives the same number $r$ (compare an arithmetic sequence, where you subtract to get a constant difference).

Geometric series (finite sum)

The sum of the first $n$ terms is

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

Both forms are equivalent. Use whichever keeps the numerator positive in your particular case.

Limiting sum (infinite series)

If $|r| < 1$ , then $r^n \to 0$ as $n \to \infty$ , so $S_n = \dfrac{a(1 - r^n)}{1 - r} \to \dfrac{a(1 - 0)}{1 - r}$ , and the series converges to

S_\infty = \frac{a}{1 - r}.

If $|r| \ge 1$ , the terms do not shrink to zero, the partial sums grow without bound (or oscillate), and the limiting sum does not exist.

Key fact

For a geometric sequence with first term $a$ and common ratio $r$ : the $n$ th term is $T_n = a r^{n-1}$ ; the sum of the first $n$ terms is $S_n = \dfrac{a(r^n - 1)}{r - 1} = \dfrac{a(1 - r^n)}{1 - r}$ (for $r \neq 1$ ); and the limiting sum exists only when $|r| < 1$ , in which case $S_\infty = \dfrac{a}{1 - r}$ . Always state the convergence test $|r| < 1$ before quoting the limiting-sum formula, because a question that says "explain why a limiting sum exists" awards a mark specifically for it.

Why an infinite sum can be finite, stage by stage

The limiting sum feels paradoxical until you watch it. Take the series $1 + \tfrac{1}{2} + \tfrac{1}{4} + \cdots$ , with $a = 1$ and $r = \tfrac{1}{2}$ , so $S_\infty = \dfrac{1}{1 - 1/2} = 2$ . Lay each term end to end on a number line: every new term is half the length of the last, so it closes half of whatever gap to $2$ remains. The partial sum keeps moving right but never passes $2$ .

Stage 1, the first term. Lay down $1$ . The running sum is $S_1 = 1$ , exactly halfway to the limit.

Stage 2, add a half. The next term is $\tfrac{1}{2}$ , closing half the gap from $1$ to $2$ . Now $S_2 = 1.5$ .

Stage 3, add a quarter. The third term $\tfrac{1}{4}$ closes half the gap again, reaching $S_3 = 1.75$ . Each step leaves exactly half the previous gap.

Stage 4, keep halving the gap. Adding $\tfrac{1}{8}$ , then $\tfrac{1}{16}$ , and so on, the sum reaches $S_5 = 1.9375$ and edges ever closer to $2$ without reaching it. The remaining gap after $n$ terms is exactly $r^n \cdot \dfrac{a}{1-r}$ , which shrinks to zero, so $S_\infty = 2$ .

Compound interest as a geometric sequence

A principal $P$ at compound rate $r$ per period produces the sequence of balances $P, P(1 + r), P(1 + r)^2, \dots$ with common ratio $1 + r$ . The balance after $n$ periods is the $(n + 1)$ th term, which gives the familiar $A = P(1 + r)^n$ .

Depreciation

An asset depreciating at rate $d$ per period has values $V_0, V_0(1 - d), V_0(1 - d)^2, \dots$ , a geometric sequence with ratio $1 - d$ . The value after $n$ periods is $V_n = V_0 (1 - d)^n$ . This is the "declining balance" method.

Repeated payments and perpetuities

A series of equal payments made at regular intervals forms a geometric sum once each payment is moved to a common time using the compound interest factor. If the payments stop after $n$ terms, use the finite sum $S_n$ (this is the future-value-of-annuity formula). If they continue forever and the per-period discount factor $v = (1 + r)^{-1}$ satisfies $|v| < 1$ (which it always does for $r > 0$ ), the limiting sum gives the present value of a perpetuity: $\text{PV} = \dfrac{\text{payment}}{r}$ .

How exam questions ask about geometric sequences and series

The context is dressed up, but each version reduces to identifying $a$ and $r$ and choosing the right formula:

"Find the value after $n$ years" of an asset that loses a fixed percentage each year. Declining-balance depreciation: $V_n = V_0(1 - d)^n$ , a geometric sequence with ratio $1 - d$ .
"Find the $n$ th term" or "which term equals $X$ ?" Use $T_n = ar^{n-1}$ ; for "which term", set it equal to $X$ and solve for $n$ with logs.
"Find the sum of the first $n$ terms" or a total of repeated equal deposits. Finite sum $S_n = \dfrac{a(r^n - 1)}{r - 1}$ .
"Find the limiting sum" or "explain why a limiting sum exists". State $|r| < 1$ first, then $S_\infty = \dfrac{a}{1 - r}$ .
"Value a scholarship / pension that pays $X forever." A perpetuity: present value $= \dfrac{X}{r}$ (the limiting sum of the discounted payments).
"Show that the balance / total is a geometric series." Write the first few terms, state $a$ and $r$ , then apply the sum formula; markers want the structure made explicit.

Worked example

Five calculations showing the geometric sequence formulas applied to a standard sequence, a depreciation problem, an alternating series, an annuity sum, and a perpetuity.

Compute $T_8$ and $S_8$ for the sequence $3, 6, 12, 24, \dots$

For the sequence $3, 6, 12, 24, \dots$ , each term is double the previous, so the first term is $a = 3$ and the common ratio is $r = 2$ .

The $n$ th term formula gives:

T_8 = a r^{7} = 3 \times 2^7 = 3 \times 128 = 384.

The finite sum formula gives:

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

Apply the depreciation model

A car worth $32000 depreciates at $20\%$ per year. Losing $20\%$ each year means multiplying by $0.8$ each year, so the successive values form a geometric sequence with $a = 32000$ and $r = 0.8$ .

After $5$ years, the value is the $(5+1)$ th term, but because we count compounding events rather than list positions, the value after $n$ years is $a r^n$ :

V = 32000 \times (0.8)^5 = 32000 \times 0.32768 \approx 10485.76.

So the car is worth about $10485.76 after $5$ years.

Check convergence then find the limiting sum

For the series $8 - 4 + 2 - 1 + \cdots$ , identify $a = 8$ and $r = -\frac{1}{2}$ .

Before computing, confirm convergence: $|r| = \frac{1}{2} < 1$ , so a limiting sum exists.

S_\infty = \frac{a}{1 - r} = \frac{8}{1 - (-\tfrac{1}{2})} = \frac{8}{\tfrac{3}{2}} = \frac{16}{3}.

Recognise an annuity as a geometric series

If you deposit $100 at the end of each year into an account paying $5\%$ per annum, each deposit grows at the factor $(1.05)$ per year. The last deposit earns no interest; the deposit made $n - 1$ years ago has compounded $n - 1$ times. Adding all contributions:

A_n = 100 + 100(1.05) + 100(1.05)^2 + \cdots + 100(1.05)^{n-1} = 100 \cdot \frac{(1.05)^n - 1}{0.05}.

This is the finite geometric series formula with $a = 100$ and $r = 1.05$ , which is the foundation of the future-value-of-annuity formula.

Value a perpetuity using the limiting sum

A scholarship pays $5000 at the end of each year forever, discounted at $4\%$ per annum. Moving each payment back to the present, the first payment is worth $5000 v$ , the second $5000 v^2$ , and so on, where $v = \frac{1}{1.04}$ .

The present value is the sum of this infinite geometric series. Since $|v| = \frac{1}{1.04} < 1$ , the limiting sum exists:

\text{PV} = \frac{5000 v}{1 - v} = \frac{5000 / 1.04}{1 - 1/1.04} = \frac{5000}{1.04 - 1} = \frac{5000}{0.04} = 125000.

So the present value of the perpetuity is $125000.

A perpetuity's value is simply the payment divided by the per-period discount rate.

Final answers: $T_8 = 384$ , $S_8 = 765$ ; car value $\approx$ $10485.76; limiting sum $= \frac{16}{3}$ ; perpetuity value $=$ $125000.

Common traps

Wrong choice of $a$: The first term $a$ is the first term of the sequence as written. In $T_n = a r^{n - 1}$ , $a = T_1$ , not $T_0$ .
Off-by-one on the exponent: $T_n$ uses $r^{n - 1}$ , not $r^n$ . The sum $S_n$ uses $r^n$ . The $n$ th term of $P, P(1 + r), P(1 + r)^2, \dots$ is $P(1 + r)^{n - 1}$ , but the compound interest balance after $n$ periods is $P(1 + r)^n$ because we count compounding events, not list positions.
Forgetting the convergence condition: The limiting sum formula needs $|r| < 1$ strictly. For $r = 1$ the sequence is constant and the partial sums diverge. For $|r| \ge 1$ but $r \neq 1$ , the partial sums grow without bound or oscillate.
Confusing depreciation rate with multiplier: A $15\%$ depreciation per year means a multiplier of $0.85$ per year, not $0.15$ . The new value is $0.85$ times the old.
Sum starting at the wrong term: Some questions list payments starting one period from now (an "ordinary annuity"), others start immediately (an "annuity due"). The first term and the number of compounded periods change accordingly.

Exam technique

State $a$ and $r$ as your first line; almost every method mark hangs on identifying them correctly. Before writing the limiting-sum formula, write the convergence test $|r| < 1$ explicitly, because "explain why a limiting sum exists" is a marked instruction. Keep depreciation as a multiplier ( $1 - d$ ), not the percentage. When a question asks "which term" or "how many years until", you will have the unknown in the exponent, so take logs and round to the next whole term if it asks when a value is first reached. For a perpetuity, quote present value $= \dfrac{\text{payment}}{r}$ and say it is the limiting sum of the discounted payments.

Exam-style questions

Questions in the style of NESA 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.

HSC-style3 marks

A machine is bought for $40000 and depreciates at

15\%

per annum. Find its value after

6

years.

Show worked answer →

Depreciation at $15\%$ per annum means the value is multiplied by $0.85$ each year. After $6$ years,

$V = 40000 (0.85)^6$ .

$(0.85)^6 \approx 0.377150$ .

$V \approx 40000 \times 0.377150 \approx 15085.98$ , i.e. $15085.98.

Markers reward the multiplier $0.85$ , the correct exponent, and an answer rounded to cents.

HSC-style3 marks

Find the limiting sum of the geometric series

1 + \frac{2}{3} + \frac{4}{9} + \frac{8}{27} + \cdots

, and explain why a limiting sum exists.

Show worked answer →

The series has first term $a = 1$ and common ratio $r = \frac{2}{3}$ .

$|r| = \frac{2}{3} < 1$ , so a limiting sum exists.

$S_\infty = \frac{a}{1 - r} = \frac{1}{1 - 2/3} = \frac{1}{1/3} = 3$ .

Markers expect identification of $a$ and $r$ , the convergence condition $|r| < 1$ , and the limiting sum formula correctly applied.

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 geometric sequence

5, 15, 45, \dots

find

T_6

and

S_6

Show worked solution →

Identify the first term and common ratio. Dividing any term by the one before, $\frac{15}{5} = 3$ , so $a = 5$ and $r = 3$ .

Apply the $n$ th-term formula.

T_6 = a r^{5} = 5 \times 3^5 = 5 \times 243 = 1215.

Apply the finite-sum formula.

S_6 = \frac{a(r^6 - 1)}{r - 1} = \frac{5(3^6 - 1)}{3 - 1} = \frac{5(729 - 1)}{2} = \frac{5 \times 728}{2} = 1820.

Marker's note: one mark for $T_6 = 1215$ from $ar^{5}$ (the exponent is $n - 1 = 5$ , not $6$ ), one for $S_6 = 1820$ from the correct sum formula. Using $r^6$ in the $n$ th term is the usual off-by-one slip.

foundation2 marks

Find the limiting sum of the geometric series

12 + 8 + \frac{16}{3} + \cdots

, first stating why it exists.

Show worked solution →

Find $a$ and $r$: The first term is $a = 12$ , and $r = \frac{8}{12} = \frac{2}{3}$ .
State the convergence test: Since $|r| = \frac{2}{3} < 1$ , a limiting sum exists.
Apply the limiting-sum formula

S_\infty = \frac{a}{1 - r} = \frac{12}{1 - \frac{2}{3}} = \frac{12}{\frac{1}{3}} = 36.

Marker's note: one mark for the convergence statement $|r| < 1$ (a marked instruction whenever a question says "why it exists"), one for $S_\infty = 36$ . Quoting the formula without the ratio test forfeits the first mark.

foundation3 marks

A machine bought for $18000 depreciates at

12\%

per annum by the declining-balance method. Find its value after

4

years, correct to the nearest cent.

Show worked solution →

Turn the depreciation rate into a multiplier. Losing $12\%$ each year means keeping $88\%$ , so the values form a geometric sequence with ratio $r = 1 - 0.12 = 0.88$ and $a = 18000$ .

Apply the declining-balance formula $V_n = V_0(1 - d)^n$ .

V_4 = 18000 \times (0.88)^4.

Evaluate the power step by step.

(0.88)^2 = 0.7744, \qquad (0.88)^4 = (0.7744)^2 = 0.59969536.

V_4 = 18000 \times 0.59969536 = 10794.52 \text{ (nearest cent).}

So the machine is worth about $10794.52 after $4$ years.

Marker's note: one mark for the multiplier $0.88$ (not $0.12$ ), one for the correct power $(0.88)^4$ , one for the value $10794.52 rounded to cents. Multiplying by $0.12$ instead of $0.88$ is the classic depreciation error.

core3 marks

An account statement lists the end-of-year balance of a savings account, in dollars, in a table. Year 0 shows 2000.00, Year 1 shows 2120.00, Year 2 shows 2247.20 and Year 3 shows 2382.03. (a) Show that the balances form a geometric sequence and state the annual interest rate. (b) Find the balance at the end of Year 10, correct to the nearest cent.

Show worked solution →

Part (a): test for a constant ratio. Divide each balance by the previous one:

\frac{2120.00}{2000.00} = 1.06, \qquad \frac{2247.20}{2120.00} = 1.06, \qquad \frac{2382.03}{2247.20} \approx 1.06.

The ratio is constant at $r = 1.06$ , so the balances form a geometric sequence. A ratio of $1.06$ is a multiplier of $1 + 0.06$ , so the annual interest rate is $6\%$ .

Part (b): use $V_n = V_0 (1.06)^n$ with $V_0 = 2000$ . The Year 10 balance is

V_{10} = 2000 \times (1.06)^{10}.

(1.06)^{10} = 1.790847\ldots,

V_{10} = 2000 \times 1.790847 = 3581.69 \text{ (nearest cent).}

So the Year 10 balance is about $3581.69.

Marker's note: one mark for showing the ratio is constant at $1.06$ , one for reading off the $6\%$ rate, one for the Year 10 balance $3581.69. Treating the yearly increases (which grow) as constant, rather than the ratio, is the trap the table is built to catch.

core3 marks

A car initially worth $4000 (in resale terms) depreciates so that each year its value is

0.85

of the year before. Find the first whole year at the end of which the value first drops below $1000.

Show worked solution →

Set up the value after $n$ years. With $a = 4000$ and $r = 0.85$ ,

V_n = 4000 (0.85)^n.

Form the inequality. We want the first $n$ with $V_n < 1000$ :

4000 (0.85)^n < 1000 \quad\Rightarrow\quad (0.85)^n < 0.25.

Take logarithms and divide (the log of $0.85$ is negative, so the inequality flips).

n \ln(0.85) < \ln(0.25) \quad\Rightarrow\quad n > \frac{\ln 0.25}{\ln 0.85} = \frac{-1.386294}{-0.162519} \approx 8.53.

Round up to the next whole year. The first integer above $8.53$ is $n = 9$ . Check: $V_9 = 4000(0.85)^9 \approx 926.47 < 1000$ , while $V_8 \approx 1089.96 > 1000$ .

So the value first drops below $1000 at the end of the $9$ th year.

Marker's note: one mark for the inequality $(0.85)^n < 0.25$ , one for taking logs and flipping the sign correctly, one for rounding up to $n = 9$ with a check. Rounding $8.53$ down to $8$ is the standard error, because year $8$ has not yet fallen below the target.

exam5 marks

At the end of each year for

20

years, Priya deposits $2000 into a fund earning

5\%

per annum compounded annually. (a) Explain why the value of the fund just after the final deposit is

A = 2000 + 2000(1.05) + 2000(1.05)^2 + \cdots + 2000(1.05)^{19}

. (b) Show that this is a geometric series and hence find its sum, correct to the nearest cent.

Show worked solution →

Part (a): track each deposit to the end. The final ( $20$ th) deposit earns no interest, contributing $2000$ . The $19$ th deposit compounds for one year, contributing $2000(1.05)$ . In general the deposit made $k$ years before the end compounds $k$ times, contributing $2000(1.05)^k$ . The very first deposit sits for $19$ years, contributing $2000(1.05)^{19}$ . Adding all $20$ contributions gives

A = 2000 + 2000(1.05) + 2000(1.05)^2 + \cdots + 2000(1.05)^{19}.

Part (b): identify $a$ and $r$ . Each term is $1.05$ times the one before, so this is a geometric series with $a = 2000$ , $r = 1.05$ and $n = 20$ terms.

A = \frac{a(r^n - 1)}{r - 1} = \frac{2000\big((1.05)^{20} - 1\big)}{1.05 - 1}.

Evaluate.

(1.05)^{20} = 2.653297705,

A = \frac{2000(2.653297705 - 1)}{0.05} = \frac{2000 \times 1.653297705}{0.05} = 2000 \times 33.065954 = 66131.91.

So the fund is worth about $66131.91 just after the final deposit.

Marker's note: one mark for explaining that the last deposit earns no interest while the first compounds $19$ times, one for writing the correct series, one for identifying $a = 2000$ , $r = 1.05$ , $n = 20$ , one for the correct substitution into $S_n$ , one for the final $66131.91. Using $n = 19$ (miscounting the terms) or $(1.05)^{20}$ in the wrong place are the common slips.

exam4 marks

A ball is dropped from a height of

10

m. After each bounce it rebounds to

\frac{3}{4}

of the height from which it fell. Assuming it keeps bouncing, find the total distance it travels before coming to rest.

Show worked solution →

Separate the initial drop from the bounces. The ball first falls $10$ m. After that, each bounce carries it up to a peak and back down the same distance, so each full bounce contributes twice its rebound height.

List the rebound heights. They form a geometric sequence with first term $10 \times \frac{3}{4} = 7.5$ and ratio $r = \frac{3}{4}$ :

7.5, \; 7.5\left(\tfrac{3}{4}\right), \; 7.5\left(\tfrac{3}{4}\right)^2, \; \dots

Sum the rebound heights with the limiting sum ( $|r| = \frac{3}{4} < 1$ ).

H = \frac{7.5}{1 - \frac{3}{4}} = \frac{7.5}{\frac{1}{4}} = 30 \text{ m}.

Combine. Each metre of rebound height is travelled twice (up then down), so the bouncing distance is $2 \times 30 = 60$ m. Adding the initial $10$ m drop,

\text{total distance} = 10 + 60 = 70 \text{ m}.

Marker's note: one mark for separating the first drop from the rebounds, one for the rebound heights as a geometric series with $a = 7.5$ , $r = \frac{3}{4}$ , one for the limiting sum $30$ m, one for doubling the rebounds and adding the initial drop to reach $70$ m. Forgetting to double the rebound heights (giving $40$ m) is the standard trap.

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

Geometric sequences

Geometric series (finite sum)

Limiting sum (infinite series)

Why an infinite sum can be finite, stage by stage

Compound interest as a geometric sequence

Depreciation

Repeated payments and perpetuities

How exam questions ask about geometric sequences and series

Compute T8T_8T8​ and S8S_8S8​ for the sequence 3,6,12,24,…3, 6, 12, 24, \dots3,6,12,24,…

Apply the depreciation model

Check convergence then find the limiting sum

Recognise an annuity as a geometric series

Value a perpetuity using the limiting sum

Exam-style questions

Practice questions

Practise this

Sources & how we know this

Compute $T_8$ and $S_8$ for the sequence $3, 6, 12, 24, \dots$