Predicting What Comes Next: Exploring Sequences and Progressions — Notes

Introduction to Sequences

A sequence is an ordered list of numbers (or other objects). Each number in the list is called a term of the sequence. The order of the terms is crucial.

Sequences help us understand patterns in various aspects of life, from nature to finance. In mathematics, they allow us to explore how numbers change and to predict future values.

We use subscript notation to denote terms in a sequence. For a sequence, the first term is denoted by $t_1$, the second term by $t_2$, the third term by $t_3$, and so on. The $n$-th term is denoted by $t_n$.

• $t_1$ = First term
• $t_2$ = Second term
• $t_n$ = $n$-th term

If we need to discuss multiple sequences, we can use different letters, like $s_1,s_2,s_3,\dots$ or $u_1,u_2,u_3,\dots$.

Sequences can be:

• Finite: They have a limited number of terms.
• Infinite: They continue without end, indicated by an ellipsis (...).

Examples of Sequences

Let's look at some common number sequences and their patterns:

1. Natural Numbers: $1,2,3,4,5,6,\dots$

• Pattern: Each term is one more than the previous term.
• $t_1=1,t_2=2,t_3=3,\dots$

2. Odd Numbers: $1,3,5,7,9,11,\dots$

• Pattern: Each term is 2 more than the previous term.
• $t_1=1,t_2=3,t_3=5,\dots$

3. Triangular Numbers: $1,3,6,10,15,21,\dots$

• Pattern: The difference between consecutive terms increases by 1 each time ($+2,+3,+4,+5,\dots$).
• These numbers can be represented as sums of natural numbers:
• $t_1=1$
• $t_2=1+2=3$
• $t_3=1+2+3=6$
• $t_4=1+2+3+4=10$
• $t_n=1+2+3+\cdots +n$

4. Square Numbers: $1,4,9,16,25,36,\dots$

• Pattern: The difference between consecutive terms increases by 2 each time ($+3,+5,+7,+9,\dots$).
• These numbers are the squares of natural numbers:
• $t_1=1^2=1$
• $t_2=2^2=4$
• $t_3=3^2=9$
• $t_n=n^2$
• They are also sums of consecutive odd numbers:
• $t_1=1$
• $t_2=1+3=4$
• $t_3=1+3+5=9$
• $t_n=1+3+5+\cdots +(2n-1)$

Other Types of Sequences

Sequences are not limited to positive integers. Terms can be:

• Fractions: For example, $1,\frac{1}{2},\frac{1}{3},\frac{1}{4},\dots$
• Negative Integers: For example, $-7,-3,1,5,9,\dots$
• Here, $s_1=-7$, and each term is 4 more than the previous term.

Example 1

Given: The sequence $1,4,7,10,13,\dots$ To find: The next four terms.

Solution: Observe the difference between consecutive terms:

The common difference is 3. Each term is 3 more than the previous term. The next term is $13+3=16$. The term after that is $16+3=19$. The term after that is $19+3=22$. The term after that is $22+3=25$.

Answer: The next four terms are 16, 19, 22, and 25.

Example 2

Given: The sequence of triangular numbers: $1,3,6,10,15,21,\dots$ To find: The terms $t_5,t_6,t_7,t_8$.

Solution: The sequence is formed by adding consecutive natural numbers:

Alternatively, notice the pattern of differences: $+2,+3,+4,+5,+6,\dots$

Answer: $t_5=15,t_6=21,t_7=28,t_8=36$.

Exercise (from reference)

Consider the sequence $1,4,7,10,13,\dots$ Can you predict the next four terms? Can you derive the first 10 terms of the sequence obtained by adding all the terms up to a given term of this sequence? (Hint: The first term is 1. The second term is $1+4=5$, the third term is $1+4+7=12$, and so on.)

Given: Sequence $1,4,7,10,13,\dots$. Let this sequence be $a_n$. Let the sequence of sums be $S_n$. $a_1=1,a_2=4,a_3=7,a_4=10,a_5=13$. The common difference is $4-1=3$. To find: The next four terms of $a_n$, and the first 10 terms of $S_n$.

Solution: Part 1: Next four terms of $a_n$ The common difference is 3.

Part 2: First 10 terms of $S_n$ $S_n$ is the sum of the first $n$ terms of $a_n$.

We need $a_{10}$. $a_{10}=a_9+3=25+3=28$.

Answer: The next four terms are 16, 19, 22, 25. The first 10 terms of the sum sequence are 1, 5, 12, 22, 35, 51, 70, 92, 117, 145.

Remember

The ellipsis (...) signifies that a sequence continues infinitely. The subscript in $t_n$ indicates the position of the term in the sequence.

Exam-smart

You might be given the first few terms of a sequence and asked to identify the pattern and predict the next terms. Alternatively, you could be asked to calculate terms based on a described rule (like "the $n$-th term is $n^2$") or to find a specific term number ($t_{100}$). You might also encounter problems where one sequence's terms are derived from another (like the sum of terms).

Explicit Rule for a Sequence

An explicit rule for a sequence, also known as an explicit formula, defines the nth term of the sequence directly in terms of its position number, n. This means you can find any term in the sequence by simply substituting its position number into the formula, without needing to know the values of preceding terms. The position number, n, is always a natural number (1, 2, 3, ...).

The notation $t_n$ (or $s_n$, $u_n$, etc.) is used to represent the nth term of a sequence. An explicit rule expresses this term as a function of n, such as $t_n=f(n)$.

Why Explicit Rules are Useful

• Direct Calculation: They allow us to calculate any term in the sequence directly, no matter how far down the sequence it is. For example, to find the 500th term, you just substitute $n=500$ into the formula.
• Independence of Previous Terms: Unlike recursive rules, explicit rules do not require you to know the value of the previous term(s). This simplifies finding terms far into the sequence.
• Membership Testing: They can be used to determine if a given number is a member of the sequence and, if so, at which position.

Example 1: The Sequence of Odd Numbers

Given: The explicit rule for the nth term of a sequence is $u_n=2n-1$.

To find: The first few terms of the sequence and the 53rd term.

Solution: To find the first few terms, substitute $n=1,2,3,\dots$ into the formula:

• For $n=1$: $u_1=2\times 1-1=2-1=1$
• For $n=2$: $u_2=2\times 2-1=4-1=3$
• For $n=3$: $u_3=2\times 3-1=6-1=5$

The first few terms are 1, 3, 5, ... which are the odd numbers.

To find the 53rd term, substitute $n=53$:

• For $n=53$: $u_{53}=2\times 53-1=106-1=105$

Answer: The first few terms are 1, 3, 5, ... The 53rd term is 105.

Using Explicit Rules to Check Membership

We can use the explicit rule to check if a number belongs to a sequence and find its position.

Given: The explicit rule $u_n=2n-1$ for the sequence of odd numbers. Check if 137 is a term and find its position.

To find: The position of the number 137 in the sequence.

Solution: Set the explicit rule equal to the number and solve for n:

• $2n-1=137$
• $2n=137+1$
• $2n=138$
• $n=\frac{138}{2}$
• $n=69$

Since $n=69$ is a natural number, 137 is a term in the sequence.

Answer: 137 is the 69th term of the odd number sequence.

Example 2: A General Explicit Formula

Given: The explicit formula $s_n=5n-2$.

To find: The first 6 terms, the 100th term, the 1000th term, and whether 308 and 471 are terms of this sequence.

Solution:

• First 6 terms:

• $s_1=5\times 1-2=3$

• $s_2=5\times 2-2=8$

• $s_3=5\times 3-2=13$

• $s_4=5\times 4-2=18$

• $s_5=5\times 5-2=23$

• $s_6=5\times 6-2=28$ The first 6 terms are 3, 8, 13, 18, 23, 28.

• 100th term:

• $s_{100}=5\times 100-2=500-2=498$

• 1000th term:

• $s_{1000}=5\times 1000-2=5000-2=4998$

• Check for 308:

• $5n-2=308$

• $5n=310$

• $n=\frac{310}{5}$

• $n=62$ Since $n=62$ is a natural number, 308 is a term.

• Check for 471:

• $5n-2=471$

• $5n=473$

• $n=\frac{473}{5}$

• $n=94.6$ Since $n=94.6$ is not a natural number, 471 is not a term of this sequence.

Answer: The first 6 terms are 3, 8, 13, 18, 23, 28. The 100th term is 498. The 1000th term is 4998. 308 is the 62nd term. 471 is not a term of this sequence.

The reason $n$ must be a natural number is that n represents the position of a term in the sequence (1st, 2nd, 3rd, etc.). Positions in a sequence are counted using positive whole numbers starting from 1. A fractional or negative position number does not make sense in this context.

Remember

An explicit rule gives you the value of the nth term ($t_n$) directly using the position number ($n$). If you solve for $n$ and it is not a natural number (1, 2, 3, ...), the number you were testing is not part of that sequence.

Exam-smart

In a case/source/data question, you might be given a sequence of numbers or an explicit formula. You could be asked to find a specific term, the first few terms, or to verify if a given number belongs to the sequence. For example, you might be given a problem about ticket prices increasing by a fixed amount for each additional person, where the explicit rule relates the total cost to the number of people. You would use the explicit rule to calculate costs for different numbers of people.

Recursive Rule for a Sequence

A recursive rule (or recursive formula) defines a sequence by relating each term to the previous term(s). To use a recursive rule, you must know the initial term(s) of the sequence.

The general form of a recursive rule for a sequence $\{t_n\}$ is:

• $t_1=\text{a specific value}$ (the first term)
• $t_n=f(t_{n-1})$ for $n\ge 2$ (a rule to find any term from the previous one)

This means that to find the second term ($t_2$), you use the first term ($t_1$) in the formula $f$. To find the third term ($t_3$), you use the second term ($t_2$) in the formula $f$, and so on.

Example 3

Given: A sequence defined by the recursive rule $u_1=1$ and $u_n=2u_{n-1}+3$ for $n\ge 2$. To find: The first four terms of the sequence and whether 133 is a term.

Solution:

• First term: $u_1=1$
• Second term ($n=2$): $u_2=2u_{2-1}+3=2u_1+3$

• Third term ($n=3$): $u_3=2u_{3-1}+3=2u_2+3$

• Fourth term ($n=4$): $u_4=2u_{4-1}+3=2u_3+3$

The first four terms are 1, 5, 13, 29.

Now, let's check if 133 is a term by continuing the sequence:

• Fifth term ($n=5$): $u_5=2u_4+3=2\times 29+3=58+3=61$
• Sixth term ($n=6$): $u_6=2u_5+3=2\times 61+3=122+3=125$
• Seventh term ($n=7$): $u_7=2u_6+3=2\times 125+3=250+3=253$

Since 133 falls between the 6th term (125) and the 7th term (253), and the terms are increasing, 133 is not a term of this sequence.

Answer: The first four terms are 1, 5, 13, 29. 133 is not a term of this sequence.

Example 4

Given: A sequence defined by the recursive rule $s_1=3$ and $s_n=s_{n-1}(s_{n-1}-1)$ for $n\ge 2$. To find: The first four terms of the sequence.

Solution:

• First term: $s_1=3$
• Second term ($n=2$): $s_2=s_{2-1}(s_{2-1}-1)=s_1(s_1-1)$

• Third term ($n=3$): $s_3=s_{3-1}(s_{3-1}-1)=s_2(s_2-1)$

• Fourth term ($n=4$): $s_4=s_{4-1}(s_{4-1}-1)=s_3(s_3-1)$

Answer: The first four terms are 3, 6, 30, 870.

A recursive rule can also involve more than just the immediately preceding term. The Virahānka–Fibonacci sequence is a famous example where each term is the sum of the two preceding terms.

The sequence is defined as:

• $V_1=1$
• $V_2=2$
• $V_n=V_{n-1}+V_{n-2}$ for $n\ge 3$

Let's find the first few terms:

• $V_1=1$
• $V_2=2$
• $V_3=V_2+V_1=2+1=3$
• $V_4=V_3+V_2=3+2=5$
• $V_5=V_4+V_3=5+3=8$

The sequence starts 1, 2, 3, 5, 8, ...

Remember

A recursive rule requires the first term(s) to be given. Without them, you cannot calculate any subsequent terms.

Exam-smart

You might be given a recursive rule and asked to find a specific term or to determine if a number belongs to the sequence. This involves step-by-step calculation using the rule.

Arithmetic Progressions

An arithmetic progression (AP) is a sequence where the difference between any two consecutive terms is constant. This constant difference is called the common difference ($d$).

• If the common difference $d$ is positive, the sequence increases.
• If the common difference $d$ is negative, the sequence decreases.

The general form of an arithmetic progression is:

Where:

• $a$ is the first term.
• $d$ is the common difference.

The explicit rule (formula for the $n$th term) for an arithmetic progression is:

Example: Arithmetic Progression

Given: The sequence 1, 5, 9, 13, ... To find: The first term, common difference, and the nth term.

Solution:

• First term ($a$): The first number in the sequence is 1. So, $a=1$.
• Common difference ($d$):
• $t_2-t_1=5-1=4$
• $t_3-t_2=9-5=4$
• $t_4-t_3=13-9=4$ The common difference is constant and is 4. So, $d=4$.
• Nth term ($t_n$): Using the formula $t_n=a+(n-1)d$:

Answer: The first term is 1, the common difference is 4, and the nth term is $t_n=4n-3$.

Example: Negative Common Difference

Given: The sequence 11, 7, 3, –1, –5, … To find: The first term and common difference.

Solution:

• First term ($a$): $a=11$.
• Common difference ($d$):
• $7-11=-4$
• $3-7=-4$
• $-1-3=-4$ The common difference is $d=-4$.

Answer: The first term is 11 and the common difference is -4.

The recursive rule for an arithmetic progression is:

• $t_1=a$
• $t_n=t_{n-1}+d$ for $n\ge 2$

This shows that to get the next term, you simply add the common difference to the current term.

Example: Recursive Rule for AP

Given: An AP with first term $a=5$ and common difference $d=-3$. To find: The recursive rule for this AP.

Solution:

• The first term is given as 5, so $t_1=5$.
• The rule to get the next term is to add the common difference ($d=-3$) to the previous term ($t_{n-1}$). So, $t_n=t_{n-1}+(-3)$, which simplifies to $t_n=t_{n-1}-3$. This rule applies for $n\ge 2$.

Answer: The recursive rule is $t_1=5$, $t_n=t_{n-1}-3$ for $n\ge 2$.

Remember

An arithmetic progression is characterized by a constant difference between consecutive terms. The explicit rule $t_n=a+(n-1)d$ and the recursive rule $t_n=t_{n-1}+d$ both describe this property.

Exam-smart

You might be given a few terms of a sequence and asked to identify if it's an AP, find its common difference, or write its $n$th term. You could also be given an explicit or recursive rule and asked to find terms or verify if it represents an AP.

Visualising an AP

When the terms of an arithmetic progression are plotted on a graph, they form a straight line.

Consider the AP from the squares pattern: 1, 5, 9, 13, 17, ... The explicit rule is $t_n=4n-3$. If we plot the (stage number, number of squares) pairs as $(n,t_n)$:

• (1, 1)
• (2, 5)
• (3, 9)
• (4, 13)
• (5, 17)

These points lie on a straight line, indicating a linear relationship between the term number ($n$) and the term value ($t_n$). This is because the explicit formula for an AP, $t_n=a+(n-1)d$, can be rearranged into the form $t_n=dn+(a-d)$, which is the equation of a straight line ($y=mx+c$, where $y=t_n$, $x=n$, $m=d$, and $c=a-d$).

Remember

Plotting the terms of an AP against their position in the sequence will always result in points that lie on a straight line.

Exam-smart

You might be shown a graph with points representing a sequence and asked to determine if it's an AP, or to find its common difference or nth term based on the visual representation. You might also be asked to predict future terms by extending the line.

Sum of the First n Natural Numbers

The sum of the first n natural numbers refers to the total obtained by adding all the whole numbers from 1 up to a given number n. This sum has a specific formula that allows us to calculate it efficiently without performing direct addition.

Deriving the Formula

Let $S$ represent the sum of the first $n$ natural numbers. We can write this as:

We can also write the same sum in reverse order:

Now, if we add these two equations together, term by term:

Notice that each pair of terms adds up to $(n+1)$:

Since there are $n$ terms in the sum, there are $n$ such $(n+1)$ values. Therefore:

Dividing by 2, we get the formula for the sum of the first $n$ natural numbers:

This formula is a fundamental result in arithmetic progressions and has historical significance, being mentioned in ancient Indian mathematical texts like Āryabhaṭa's Āryabhaṭīya.

Example 1: Sum of the first 10 natural numbers

Given: We want to find the sum of the first 10 natural numbers. To find: The sum, $S_{10}$.

Solution: Using the formula $S_n=\frac{n(n+1)}{2}$ with $n=10$:

Answer: The sum of the first 10 natural numbers is 55.

Example 2: Sum of the first 100 natural numbers

Given: We want to find the sum of the first 100 natural numbers. To find: The sum, $S_{100}$.

Solution: Using the formula $S_n=\frac{n(n+1)}{2}$ with $n=100$:

Answer: The sum of the first 100 natural numbers is 5050.

Using the Formula for Consecutive Numbers

The formula $S_n=\frac{n(n+1)}{2}$ can also be used to find the sum of a series of consecutive natural numbers that do not start from 1.

To find the sum of natural numbers from $a$ to $b$ (inclusive), where $a>1$, we can use the principle of inclusion-exclusion: Sum(a to b) = Sum(1 to b) - Sum(1 to (a-1)) Sum(a to b) = $S_b-S_{a-1}$ Sum(a to b) = $\frac{b(b+1)}{2}-\frac{(a-1)((a-1)+1)}{2}$ Sum(a to b) = $\frac{b(b+1)}{2}-\frac{(a-1)a}{2}$

Example 3: Sum of numbers from 25 to 58

Given: The series of numbers 25, 26, 27, ..., 58. To find: The sum of this series.

Solution: We need to find the sum from 25 to 58. This can be calculated as (Sum of first 58 natural numbers) - (Sum of first 24 natural numbers).

Sum(25 to 58) = $S_{58}-S_{24}=1711-300=1411$.

Answer: The sum of the numbers from 25 to 58 is 1411.

Connection to Triangular Numbers

The sequence of triangular numbers (1, 3, 6, 10, 15, ...) represents the number of dots required to form equilateral triangles of increasing size. The $n^{th}$ triangular number, denoted as $t_n$, is precisely the sum of the first $n$ natural numbers.

Thus, the formula for the $n^{th}$ triangular number is:

Example 4: Finding triangular numbers

Given: The formula for the $n^{th}$ triangular number is $t_n=\frac{n(n+1)}{2}$. To find: The 10th, 17th, and 80th triangular numbers.

Solution: For the 10th triangular number ($n=10$):

For the 17th triangular number ($n=17$):

For the 80th triangular number ($n=80$):

Answer: The 10th triangular number is 55, the 17th is 153, and the 80th is 3240.

Remember

The formula $S_n=\frac{n(n+1)}{2}$ is for the sum of the first n natural numbers (starting from 1). For sums of consecutive numbers not starting from 1, subtract the sum of the numbers before the starting number from the sum up to the ending number.

Exam-smart

You might be given a word problem asking for the total number of items arranged in a pattern where each subsequent row/group has one more item than the previous (like marbles in rows, or days of a challenge). This is directly related to finding the sum of natural numbers. For example, if a gardener plants saplings in rows, with the first row having 1, the second having 2, and so on, up to the 20th row, the total number of saplings would be $S_{20}$. You'll need to identify 'n' from the problem description.

Geometric Progressions

A Geometric Progression (GP) is a sequence of numbers where each term after the first is found by multiplying the previous one by a fixed, non-zero number called the common ratio.

Consider the sequence: 3, 6, 12, 24, ... Here, the first term ($a$) is 3. To get the second term, we multiply the first by 2: $3\times 2=6$. To get the third term, we multiply the second by 2: $6\times 2=12$. To get the fourth term, we multiply the third by 2: $12\times 2=24$. The common ratio ($r$) is 2.

The general form of a GP is: $a,ar,a{r}^2,a{r}^3,\dots ,a{r}^{n-1},\dots$ where:

• $a$ is the first term.
• $r$ is the common ratio.
• $a{r}^{n-1}$ is the nth term of the GP.

Example 6: Is 1, 2, 4, 8, 16, … a geometric progression? If so, what is the common ratio?

Given: The sequence 1, 2, 4, 8, 16, … To find: If it is a GP and its common ratio.

Solution: Calculate the ratio between consecutive terms:

Since the ratio is constant, the sequence is a GP. The common ratio ($r$) is 2.

Answer: Yes, it is a geometric progression with a common ratio of 2.

Example 7: Is 1, 3, 9, 27, 81, … a geometric progression? If so, what is the common ratio?

Given: The sequence 1, 3, 9, 27, 81, … To find: If it is a GP and its common ratio.

Solution: Calculate the ratio between consecutive terms:

The ratio is constant. The common ratio ($r$) is 3.

Answer: Yes, it is a geometric progression with a common ratio of 3.

Example 8: Is 1, –1, 1, –1, 1, … a geometric progression? If so, what is the common ratio?

Given: The sequence 1, –1, 1, –1, 1, … To find: If it is a GP and its common ratio.

Solution: Calculate the ratio between consecutive terms:

The ratio is constant. The common ratio ($r$) is -1.

Answer: Yes, it is a geometric progression with a common ratio of -1.

Example 9: Check whether the sequence 5, $\frac{15}{4}$, $\frac{45}{16}$, $\frac{135}{64}$, … is a geometric progression and find its nth term.

Given: The sequence 5, $\frac{15}{4}$, $\frac{45}{16}$, $\frac{135}{64}$, … To find: If it is a GP, its common ratio, and the nth term.

Solution: The first term ($a$) is 5. Calculate the ratio of consecutive terms: Term 2 / Term 1 = $\frac{15\mathrm{/}4}{5}=\frac{15}{4\times 5}=\frac{15}{20}=\frac{3}{4}$ Term 3 / Term 2 = $\frac{45\mathrm{/}16}{15\mathrm{/}4}=\frac{45}{16}\times \frac{4}{15}=\frac{45\times 4}{16\times 15}=\frac{3\times 15\times 4}{4\times 4\times 15}=\frac{3}{4}$ Term 4 / Term 3 = $\frac{135\mathrm{/}64}{45\mathrm{/}16}=\frac{135}{64}\times \frac{16}{45}=\frac{135\times 16}{64\times 45}=\frac{3\times 45\times 16}{4\times 16\times 45}=\frac{3}{4}$ The common ratio ($r$) is $\frac{3}{4}$. The nth term of a GP is given by $t_n=a{r}^{n-1}$. Substituting $a=5$ and $r=\frac{3}{4}$:

Answer: Yes, the sequence is a geometric progression with first term $a=5$ and common ratio $r=\frac{3}{4}$. The nth term is $5\times {\left(\frac{3}{4}\right)}^{n-1}$.

Fun with Fractals

Fractals are self-similar patterns that repeat at different scales. This means that if you zoom in on a part of a fractal, it looks like the whole shape. Fractals are found in nature, such as in tree branches or snowflakes.

The Sierpiński Triangle: This fractal is created by starting with an equilateral triangle and repeatedly removing the central triangle formed by joining the midpoints of its sides.

• Stage 0: 1 black triangle (the initial triangle).
• Stage 1: The initial triangle is divided into 4 smaller triangles, and the central one is removed, leaving 3 black triangles.
• Stage 2: Each of the 3 black triangles from Stage 1 is further divided and has its central part removed, resulting in $3\times 3=9$ black triangles.
• Stage 3: Each of the 9 black triangles from Stage 2 is processed, resulting in $9\times 3=27$ black triangles.

The number of black triangles at each stage forms a sequence: 1, 3, 9, 27, ... This sequence is a Geometric Progression with:

• First term ($a$) = 1
• Common ratio ($r$) = 3 (since each stage triples the number of triangles)

The number of black triangles at the nth stage is given by $t_n=a{r}^{n-1}=1\times 3^{n-1}=3^{n-1}$. However, if Stage 0 is considered $n=0$, the formula is $3^n$. Let's stick to the textbook's notation where Stage 0 has $3^0=1$ black triangles, Stage 1 has $3^1=3$, Stage 2 has $3^2=9$, and Stage 3 has $3^3=27$. So, the number of black triangles at stage $n$ is $3^n$.

Area of the black region in Sierpiński Triangle:

• If the area of the triangle in Stage 0 is 1 square unit.
• In Stage 1, the triangle is divided into 4 equal parts, and one is removed, so $\frac{3}{4}$ of the area remains black. Area = $1\times \frac{3}{4}=\frac{3}{4}$.
• In Stage 2, each of the 3 black triangles from Stage 1 has $\frac{3}{4}$ of its area remaining. So, the total black area is $\frac{3}{4}\times \frac{3}{4}={\left(\frac{3}{4}\right)}^2$.
• In Stage 3, the black area is ${\left(\frac{3}{4}\right)}^3$.

The sequence of areas is 1, $\frac{3}{4}$, ${\left(\frac{3}{4}\right)}^2$, ${\left(\frac{3}{4}\right)}^3$, ... This is a Geometric Progression with:

• First term ($a$) = 1
• Common ratio ($r$) = $\frac{3}{4}$

The area of the black region at stage $n$ is given by $s_n=a{r}^{n-1}=1\times {\left(\frac{3}{4}\right)}^{n-1}$. Again, if Stage 0 is $n=0$, the formula is ${\left(\frac{3}{4}\right)}^n$. Following the table, Stage 0 has area $(\frac{3}{4}{)}^0=1$, Stage 1 has area $(\frac{3}{4}{)}^1=\frac{3}{4}$, Stage 2 has area $(\frac{3}{4}{)}^2$, and Stage 3 has area $(\frac{3}{4}{)}^3$. So, the area of the black region at stage $n$ is ${\left(\frac{3}{4}\right)}^n$.

Example: Fractal Properties

Given: The Sierpiński triangle construction. To find: The rule for the number of black triangles and the area of the black region at stage $n$.

Solution: Number of black triangles: Stage 0: 1 triangle ($3^0$) Stage 1: 3 triangles ($3^1$) Stage 2: 9 triangles ($3^2$) Stage 3: 27 triangles ($3^3$) The number of black triangles at stage $n$ is $3^n$. This is a GP with $a=1$ and $r=3$.

Area of the black region: Stage 0: Area = 1 ($(\frac{3}{4}{)}^0$) Stage 1: Area = $\frac{3}{4}$ ($(\frac{3}{4}{)}^1$) Stage 2: Area = $\frac{9}{16}$ ($(\frac{3}{4}{)}^2$) Stage 3: Area = $\frac{27}{64}$ ($(\frac{3}{4}{)}^3$) The area of the black region at stage $n$ is ${\left(\frac{3}{4}\right)}^n$. This is a GP with $a=1$ and $r=\frac{3}{4}$.

Answer: The number of black triangles at stage $n$ is $3^n$. The area of the black region at stage $n$ is ${\left(\frac{3}{4}\right)}^n$.

Visualising a GP

When we plot the terms of a Geometric Progression against their position number, the points do not lie on a straight line (unless $r=1$ or $a=0$, which are trivial cases). Instead, they show a rapid increase or decrease, characteristic of exponential growth or decay.

Fig. 8.9 shows the points (1, 3), (2, 6), (3, 12), (4, 24), (5, 48) from the growing pattern of squares (a GP with $a=3,r=2$). The points curve upwards, showing rapid growth.

Fig. 8.10 (A) shows points related to the Sierpiński triangle (number of black triangles: 1, 3, 9, 27, 81,...). These points also show rapid growth.

Fig. 8.10 (B) shows points related to the Sierpiński triangle (area of black region: 1, $\frac{3}{4}$, $\frac{9}{16}$, $\frac{27}{64}$,... ). These points show a decrease, approaching zero.

Example 10: A ball is dropped from a height of 24 feet. Each time it bounces up to $\frac{3}{4}$ of its previous height.

Given: Initial height = 24 feet, bounce height ratio = $\frac{3}{4}$. To find: (a) Sequence of heights for five bounces. (b) Number of bounces for the ball to remain below $\frac{1}{4}$ of the original height.

Solution: (a) Sequence of heights after each bounce: This is a GP where the first term ($a$) is the height after the first bounce. First bounce height: $24\times \frac{3}{4}=18$ feet. So, $a=18$. The common ratio ($r$) is $\frac{3}{4}$. The sequence of heights for five successive bounces is:

$t_2=18\times \frac{3}{4}=\frac{54}{4}=13.5$ feet $t_3=13.5\times \frac{3}{4}=\frac{27}{2}\times \frac{3}{4}=\frac{81}{8}=10.125$ feet $t_4=10.125\times \frac{3}{4}=\frac{81}{8}\times \frac{3}{4}=\frac{243}{32}=7.59375$ feet $t_5=7.59375\times \frac{3}{4}=\frac{243}{32}\times \frac{3}{4}=\frac{729}{128}=5.6953125$ feet

(b) Number of bounces for the ball to remain below $\frac{1}{4}$ of the original height. Original height = 24 feet. Target height = $\frac{1}{4}\times 24=6$ feet. We need to find $n$ such that the height after the $n$-th bounce ($t_n$) is less than 6 feet. The formula for the height after the $n$-th bounce is $t_n=a{r}^{n-1}=18\times {\left(\frac{3}{4}\right)}^{n-1}$. We need $18\times {\left(\frac{3}{4}\right)}^{n-1}<6$. Divide by 18: ${\left(\frac{3}{4}\right)}^{n-1}<\frac{6}{18}=\frac{1}{3}$.

Let's check the heights we calculated: $t_1=18$ (not < 6) $t_2=13.5$ (not < 6) $t_3=10.125$ (not < 6) $t_4=7.59375$ (not < 6) $t_5=5.6953125$ (which is < 6)

So, after the 5th bounce, the ball's height is below 6 feet.

Let's calculate the height for the 6th bounce to be sure: $t_6=t_5\times \frac{3}{4}=5.6953125\times \frac{3}{4}=\frac{729}{128}\times \frac{3}{4}=\frac{2187}{512}\approx 4.27$ feet. This is also below 6 feet.

The question asks "How many bounces are required for the ball to remain below...". This means we are looking for the first bounce where the subsequent height is below the target. After the 5th bounce, the height reached is $5.6953125$ ft, which is less than 6 ft. So, 5 bounces are required for the ball to remain below that height.

Answer: (a) The sequence of heights for five successive bounces is 18, 13.5, 10.125, 7.59375, 5.6953125 feet. (b) 5 bounces are required for the ball to remain below $\frac{1}{4}$ of its original height.

Remember

In a GP, the ratio of any term to its preceding term is constant. This constant is the common ratio (r). If $\mathrm{\mid }r\mathrm{\mid }<1$, the terms get smaller. If $\mathrm{\mid }r\mathrm{\mid }>1$, the terms get larger. If $r=1$, all terms are the same. If $r=-1$, terms alternate in sign.

Exam-smart

Questions involving GPs often give you two terms and ask for the first term or common ratio, or ask you to find a specific term. You might also see problems on bouncing balls or compound interest which model GPs. Look for a constant multiplier between successive values.