Applications of APs and GPs Revision Notes for HSC SSCE Mathematics Advanced

Applications of APs and GPs

Introduction

This topic reviews and applies the key results about arithmetic progressions (APs) and geometric progressions (GPs). These sequences appear frequently in finance, population modeling, and many real-world situations where quantities grow or decline in predictable patterns.

infoNote

Key terminology:

A sequence is an ordered list of numbers
A series is the sum of the terms in a sequence
For example: $1, 3, 5, \ldots$ is a sequence, while $1 + 3 + 5 + \ldots$ is a series

Arithmetic sequences

An arithmetic sequence (or AP) is a sequence where the difference between consecutive terms stays constant. This constant value is called the common difference, denoted by $d$ .

Key properties of arithmetic sequences

chatImportant

Definition: A sequence $T_n$ is an arithmetic sequence if:

$T_n - T_{n-1} = d \text{ for all } n \geq 2$

where $d$ is the common difference.

The nth term formula:

$T_n = a + (n-1)d$

where:

$a$ is the first term
$d$ is the common difference
$n$ is the term number

infoNote

Three numbers in AP: Three numbers $a$ , $x$ , and $b$ are in AP if:

$x = \frac{a + b}{2}$

In other words, the middle term is the average of the other two terms.

Sum formulas: The sum $S_n$ of the first $n$ terms can be calculated using either formula:

$S_n = \frac{1}{2}n(a + \ell)$

Use this formula when you know the last term $\ell = T_n$ .

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

Use this formula when you know the common difference $d$ .

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Worked Example: Salaries and APs

Problem: Georgia earned $50,000 in her first year at Information Holdings, and her salary increased by $6,000 every year. She worked at the company for 12 years.

a) What was her annual salary in her final year?

b) What were her total earnings over the 12 years?

Solution:

Her annual salaries form the series: $50000 + 56000 + 62000 + \ldots$ with 12 terms.

This is an AP with $a = 50000$ , $d = 6000$ , and $n = 12$ .

Part a: Her final salary is the 12th term $T_{12}$ .

Final salary $= a + 11d$

$= 50000 + 11 \times 6000$

$= 50000 + 66000$

$= \$ 116000$

Part b: Her total earnings are the sum $S_{12}$ of the first 12 terms.

Using the first sum formula:

Total earnings $= \frac{1}{2}n(a + \ell)$

$= \frac{1}{2} \times 12 \times (50000 + 116000)$

$= 6 \times 166000$

$= \$996000$

Alternatively, using the second sum formula:

Total earnings $= \frac{1}{2}n(2a + (n-1)d)$

$= \frac{1}{2} \times 12 \times (2 \times 50000 + 11 \times 6000)$

$= 6 \times (100000 + 66000)$

$= \$996000$

infoNote

Exam tip: Both sum formulas give the same answer. Choose the one that matches the information you have.

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Worked Example: Ticket Pricing

Problem: The Roxanne Cinema has a group concession. It charges $24 for the first ticket and $16 for each additional ticket.

a) How much would 20 tickets cost?

b) Find a formula for the cost of $n$ tickets.

c) How many people are in a group whose tickets cost $600?

Solution:

The costs form the sequence: $24, $40, $56, $\ldots$

This is an AP with $a = 24$ and $d = 16$ .

Part a: The cost of 20 tickets is the 20th term $T_{20}$ .

Cost of 20 tickets $= a + 19d$

$= 24 + 19 \times 16$

$= 24 + 304$

$= \$328$

Part b: Using the nth term formula:

Cost of $n$ tickets $= a + (n-1)d$

$= 24 + (n-1) \times 16$

$= 24 + 16n - 16$

$= 16n + 8$ dollars

Part c: Set the cost equal to 600 and solve for $n$ :

$16n + 8 = 600$

$16n = 592$

$n = 37$ tickets

Problems involving specific years require careful counting. It helps to think in terms of "years after" a starting point.

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Worked Example: Counting with Named Years

Problem: Gulgarindi Council had 2870 complaints in 2006, but only 2170 in 2016. The number of complaints decreased by the same amount each year.

a) What was the total number of complaints during these years?

b) By how much did the complaints decrease each year?

c) How many complaints were there in 2008?

d) Find a formula for the number of complaints in the nth year.

e) If the trend continued, in what year would there be no complaints?

Solution:

Important: The first year (2006) is year 1, the second year (2007) is year 2, and so on. The 11th year is 2016.

In general, the nth year of the problem is the nth year after 2005.

The successive complaints form an AP with $a = 2870$ , $\ell = 2170$ , and $n = 11$ .

Part a: Total complaints is the sum $S_{11}$ :

Total complaints $= \frac{1}{2} \times 11 \times (a + \ell)$

$= \frac{1}{2} \times 11 \times (2870 + 2170)$

$= \frac{1}{2} \times 11 \times 5040$

$= 27720$

Part b: We need to find the common difference $d$ (which will be negative).

Using $T_{11} = 2170$ :

$a + 10d = 2170$

$2870 + 10d = 2170$

$10d = -700$

$d = -70$

The complaints decreased by 70 each year.

Part c: The year 2008 is the third year, so we find $T_3$ :

Complaints in 2008 $= a + 2d$

$= 2870 + 2 \times (-70)$

$= 2870 - 140$

$= 2730$

Part d: The number of complaints in year $n$ is:

$T_n = a + (n-1)d$

$= 2870 + (n-1) \times (-70)$

$= 2870 - 70n + 70$

$= 2940 - 70n$

Part e: Set $T_n = 0$ and solve:

$2940 - 70n = 0$

$70n = 2940$

$n = 42$

The year would be $2005 + 42 = 2047$ .

chatImportant

Exam tip: When working with named years, always clearly define which year is "year 1" of your sequence to avoid counting errors.

Geometric sequences

A geometric sequence (or GP) is a sequence where the ratio between consecutive terms stays constant. This constant value is called the common ratio, denoted by $r$ .

Key properties of geometric sequences

chatImportant

Definition: A sequence $T_n$ is a geometric sequence if:

$\frac{T_n}{T_{n-1}} = r \text{ for all } n \geq 2$

where $r$ is the common ratio.

Important: Neither the ratio $r$ nor any term of a GP can be zero.

The nth term formula:

$T_n = ar^{n-1}$

where:

$a$ is the first term
$r$ is the common ratio
$n$ is the term number

infoNote

Three numbers in GP: Three numbers $a$ , $x$ , and $b$ are in GP if:

$x^2 = ab$

In other words, the middle term squared equals the product of the other two terms.

Sum formulas: The sum $S_n$ of the first $n$ terms depends on whether $r$ is greater or less than 1:

$S_n = \frac{a(r^n - 1)}{r - 1} \text{ (easier when } r > 1\text{)}$

$S_n = \frac{a(1 - r^n)}{1 - r} \text{ (easier when } r < 1\text{)}$

chatImportant

Limiting sum: The limiting sum $S_{\infty}$ exists if and only if $|r| < 1$ (that is, $-1 < r < 1$ ). When it exists:

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

This represents the sum of all terms as $n$ approaches infinity.

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Worked Example: Geometric Sequence with $r > 1$

Problem: The town of Elgin grew rapidly after a new distillery opened. In the first year afterwards, there were 15 car accidents, but the number doubled every year for eight years.

a) Find the number of accidents in the eighth year.

b) Find the total number of accidents over these eight years.

c) What percentage of total accidents occurred in the final year?

Solution:

The number of accidents per year forms the sequence: $15, 30, 60, \ldots$

This is a GP with $a = 15$ and $r = 2$ .

Part a: The number of accidents in year 8 is $T_8$ :

Number of accidents $= ar^7$

$= 15 \times 2^7$

$= 15 \times 128$

$= 1920$

Part b: The total accidents over 8 years is $S_8$ :

Since $r > 1$ , we use the first sum formula:

Total accidents $= \frac{a(r^8 - 1)}{r - 1}$

$= \frac{15 \times (2^8 - 1)}{2 - 1}$

$= \frac{15 \times (256 - 1)}{1}$

$= 15 \times 255$

$= 3825$

Part c: To find the percentage:

$\frac{\text{Accidents in final year}}{\text{Total accidents}} = \frac{1920}{3825} \approx 50.20\%$

Key observation: In a GP with $r > 1$ , the later terms dominate the sum. Here, the final year alone accounts for about half of all accidents.

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Worked Example: Geometric Sequence with $r < 1$

Problem: Sales from the Gumnut Softdrinks Factory are declining by 6% every year. In 2016, 50,000 bottles were sold.

a) How many bottles will be sold in 2025?

b) How many bottles will be sold in total during 2016–2025?

Solution:

Here 2016 is year 1, 2017 is year 2, and so on, making 2025 year 10.

If sales decline by 6%, then each year's sales are 94% of the previous year.

The annual sales form a GP with $a = 50000$ and $r = 0.94$ .

Part a: Sales in 2025 are the 10th term $T_{10}$ :

Sales in 2025 $= ar^9$

$= 50000 \times 0.94^9$

$\approx 28650$ bottles

Part b: Total sales over 10 years is $S_{10}$ :

Since $r < 1$ , we use the second sum formula:

Total sales $= \frac{a(1 - r^{10})}{1 - r}$

$= \frac{50000 \times (1 - 0.94^{10})}{0.06}$

$\approx 384487$ bottles

Limiting sums

When the common ratio of a GP satisfies $|r| < 1$ , the sum $S_n$ converges to a finite limit as $n \to \infty$ . This allows us to discuss the "eventual" sum of all terms.

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Worked Example: Limiting Sum Application

Problem: Consider again the Gumnut Softdrinks Factory where sales decline by 6% annually and 50,000 bottles were sold in 2016. Suppose the company continues indefinitely.

a) What would the total sales from 2016 onwards eventually be?

b) What proportion of those sales would occur by the end of 2025?

Solution:

The sales form a GP with $a = 50000$ and $r = 0.94$ .

Since $-1 < 0.94 < 1$ , the limiting sum exists.

Part a: The eventual sales are:

$S_{\infty} = \frac{a}{1 - r} = \frac{50000}{1 - 0.94} = \frac{50000}{0.06} \approx 833333 \text{ bottles}$

Part b: From the previous example, we found $S_{10} \approx 384487$ .

The proportion is:

$\frac{S_{10}}{S_{\infty}} = \frac{50000 \times (1 - 0.94^{10})}{0.06} \times \frac{0.06}{50000} = 1 - 0.94^{10} \approx 46.14\%$

Key insight: Even though sales continue indefinitely, almost half of all eventual sales occur in the first 10 years.

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Worked Example: Trigonometric Application

Problem: Consider the series $1 - \tan^2 x + \tan^4 x - \ldots$ where $x$ is an acute angle.

a) For what values of $x$ does the series have a limiting sum?

b) What is this limiting sum when it exists?

Solution:

Part a: This is a GP with $a = 1$ and $r = -\tan^2 x$ .

The limiting sum exists when $-1 < r < 1$ , which means:

$-1 < -\tan^2 x < 1$

Since $\tan^2 x \geq 0$ , we need:

$-\tan^2 x < 1$

$-1 < \tan^2 x$

This gives us: $\tan^2 x < 1$

Therefore: $-1 < \tan x < 1$

Since $\tan 0° = 0$ and $\tan 45° = 1$ , and $x$ is acute:

$0° < x < 45°$

Part b: When the series converges:

$S_{\infty} = \frac{a}{1 - r} = \frac{1}{1 - (-\tan^2 x)} = \frac{1}{1 + \tan^2 x}$

Using the identity $1 + \tan^2 x = \sec^2 x$ :

$= \frac{1}{\sec^2 x} = \cos^2 x$

bookmarkSummary

Key Points to Remember:

Arithmetic sequences have a constant difference between terms. Use them for linear growth situations.
Geometric sequences have a constant ratio between terms. Use them for exponential growth or decay.
For sum formulas: Choose based on what information you have:
- For APs: Use $S_n = \frac{1}{2}n(a + \ell)$ when you know the last term
- For GPs: Use $\frac{r^n - 1}{r - 1}$ when $r > 1$ , and $\frac{1 - r^n}{1 - r}$ when $r < 1$
Limiting sums only exist for GPs when $|r| < 1$ . Use $S_{\infty} = \frac{a}{1-r}$ to find them.
When working with named years, clearly define which calendar year corresponds to $n = 1$ in your sequence.

Applications of APs and GPs (HSC SSCE Mathematics Advanced): Revision Notes