Linear and Quadratic Sequences

Create a lesson plan using the following information - 1. Linear or Arithmetic Sequences

A linear sequence, also known as an arithmetic sequence, is a sequence in which each term is found by adding a constant number to the previous term. This constant number that is added is called the common difference.

Some examples are:

1, 4, 7, 10 … (where 3 is added to the previous term each time)

10, 8, 6, 4 … (where –2 is added to the previous term each time).

Worked Example 1

Which of the following are arithmetic sequences?

1. 30, 42, 61 …
2. 30, 15, 5 …
3. 10, 30, 50 ...

Continued Solutions to Worked Example 1:

Start with: 30, 42, 61 … 𝑇2 – 𝑇1 = 42 – 30 = 12 𝑇3 – 𝑇2 = 61 – 42 = 9 Since 𝑇2 – 𝑇1 ≠ 𝑇3 – 𝑇2, the sequence is not arithmetic.

Start with: 30, 15, 5 … 𝑇2 – 𝑇1 = 15 – 30 = –15 𝑇3 – 𝑇2 = 5 – 15 = –10 Since 𝑇2 – 𝑇1 ≠ 𝑇3 – 𝑇2, the sequence is not arithmetic.

Start with: 10, 30, 50 ... 𝑇2 – 𝑇1 = 30 – 10 = 20 𝑇3 – 𝑇2 = 50 – 30 = 20 Since 𝑇2 – 𝑇1 = 𝑇3 – 𝑇2, the sequence is arithmetic. 2. Using the Formula for the 𝑛th Term of a Linear Sequence

Each arithmetic sequence has a unique formula (known as an 𝑛th term formula) that describes that sequence. It is always written in the following way: 𝑇𝑛 = 𝑎𝑛 + 𝑏, where 𝑎 and 𝑏 are constants, and 𝑛 represents the position of the term. This formula can be used to find any term in the sequence by substituting a particular value for 𝑛.

Worked Example 2

If, for a particular sequence, 𝑇𝑛 = 2𝑛 – 1, find:

1. The values of the first 3 terms of the sequence
2. The value of the 200th term in the sequence.

Continued Solutions to Worked Example 2:

1. Substitute 𝑛 = 1 into the formula: 𝑇1 = 2(1) – 1 = 1 Substitute 𝑛 = 2 into the formula: 𝑇2 = 2(2) – 1 = 3 Substitute 𝑛 = 3 into the formula: 𝑇3 = 2(3) – 1 = 5

2. Substitute 𝑛 = 200 into the formula: 𝑇200 = 2(200) – 1 = 399

Worked Example 3

Find the value of the fifth term of the sequence with 𝑇𝑛 = –7 – 3𝑛

Continued Solutions to Worked Example 3:

Substitute 𝑛 = 5 into the formula: 𝑇5 = –7 – 3(5) = –22

3. Finding the Formula of the 𝑛th Term of a Linear Sequence (Position-to-term rule)

When trying to find the 𝑛th term formula of an arithmetic sequence, take note of the following:

The 𝑛th term formula will always have the form 𝑇𝑛 = 𝑎𝑛 + 𝑏, where 𝑎 and 𝑏 are constants, and 𝑛 represents the position of the term.

The value 𝑎 is always the common difference.

Find the value of 𝑏 by substituting any 𝑛 value.

To find the formula for the 𝑛th term of a linear (arithmetic) sequence, follow these steps:

Identify the common difference (𝑑): In an arithmetic sequence, the common difference (𝑑) is the constant value that is added or subtracted to each term to get to the next term. Look for a consistent difference between consecutive terms.

Determine the first term (𝑎1): Find the value of the first term in the sequence (𝑎1). It is usually given or can be identified from the sequence.

Write the general formula: The formula for the 𝑛th term of an arithmetic sequence is given by:

𝑎𝑛 = 𝑎1 + 𝑑(𝑛 - 1)

In this formula:

𝑎n represents the 𝑛th term in the sequence. 𝑎1 represents the first term in the sequence. 𝑛 represents the position of the term in the sequence. 𝑑 represents the common difference. Substitute the values: Plug in the values of 𝑎1 and 𝑑 into the general formula to get the specific formula for the 𝑛th term.

Example: Let's say we have an arithmetic sequence with a first term of 2 and a common difference of 3. We want to find the formula for the 𝑛th term.

Using the general formula: 𝑎𝑛 = 𝑎1 + 𝑑(𝑛 - 1)

Substitute the values: 𝑎𝑛 = 2 + 3(𝑛 - 1)

Simplify the equation: 𝑎𝑛 = 2 + 3𝑛 - 3 𝑎𝑛 = 3𝑛 - 1

Therefore, the formula for the 𝑛th term of this arithmetic sequence is 𝑎𝑛 = 3𝑛 - 1.

Worked Example 4

Consider the arithmetic sequence: 7, 12, 17, 22, 27. We want to find the formula for the 𝑛th term.

Continued Solutions to Worked Example 4:

If we follow the steps we will start by identifying the common difference (𝑑): Look for the consistent difference between consecutive terms. In this sequence, the common difference is 5, as each term is obtained by adding 5 to the previous term.

Next, we determine the first term (𝑎1): Identify the value of the first term in the sequence. In this case, the first term is 7.

We now write the general formula: The formula for the 𝑛th term of an arithmetic sequence is given by: 𝑎𝑛 = 𝑎1 + 𝑑(𝑛 - 1)

Substitute the values: Plug in the values of 𝑎1 and 𝑑 into the general formula. 𝑎𝑛 = 7 + 5(𝑛 - 1)

Simplify the equation: 𝑎𝑛 = 7 + 5𝑛 - 5 𝑎𝑛 = 5𝑛 + 2

Therefore, the formula for the 𝑛th term of this arithmetic sequence is 𝑎𝑛 = 5𝑛 + 2. This means that if we want to find the value of the 𝑛th term in this sequence, we can substitute the value of 𝑛 into the formula 𝑎𝑛 = 5𝑛 + 2. For example, if we want to find the 8th term, we substitute 𝑛 = 8 into the formula: 𝑎8 = 5(8) + 2 = 42.

Worked Example 5

Work out the 𝑛th term formula for each of these sequences:

1. 5, 12, 19 …
2. 25, 22, 19 ...

Continued Solutions to Worked Example 5:

1. 5, 12, 19 … Calculate the common difference: 𝑎 = 𝑇2 – 𝑇1 = 12 – 5 = 7 or 𝑎 = 𝑇3 – 𝑇2 = 19 – 12 = 7

Substitute 𝑎 = 7 into the formula 𝑇𝑛 = 𝑎𝑛 + 𝑏: ∴ 𝑇𝑛 = 7𝑛 + 𝑏

Substitute the position and corresponding value of any term into the formula: Since : 𝑇1 = 5 ∴ 5 = 7(1) + 𝑏 ∴ 𝑏 = –2

Now substitute the 𝑏 value into the equation: ∴ 𝑇𝑛 = 7𝑛 – 2

2. 25, 22, 19 ... Calculate the common difference: 𝑎 = 𝑇2 – 𝑇1 = 22 – 25 = –3 or 𝑎 = 𝑇3 – 𝑇2 = 19 – 22 = –3

Substitute 𝑎 = –3 into the formula 𝑇𝑛 = 𝑎𝑛 + 𝑏: ∴ 𝑇𝑛 = -3𝑛 + 𝑏

Substitute the position and corresponding value of any term into the formula: Since: 𝑇1 = 25 ∴ 25 = –3(1) + 𝑏 ∴ 𝑏 = 28

Now substitute the 𝑏 value into the equation: ∴ 𝑇𝑛 = –3𝑛 + 28 This could also be written as 𝑇𝑛 = 28 – 3𝑛

Worked Example 6

Is 157 a term in the sequence 𝑇𝑛 = 2𝑛 + 13?

Continued Solutions to Worked Example 6:

Start with: 𝑇𝑛 = 2𝑛 + 13

Substitute: 𝑇𝑛 = 157 ∴ 157 = 2𝑛 + 13

Solve for 𝑛: 𝑛 = 72

Since 72 is an integer, and it is possible to have a 72 term in the sequence, we can say that 157 is a term in the sequence.

Worked Example 7

Is 105 a term in the sequence 3, 7, 11 …?

Continued Solutions to Worked Example 7:

First, find the formula for the 𝑛th term:

Start by finding the common difference: 𝑎 = 𝑇2 – 𝑇1 = 7 – 3 = 4 or 𝑎 = 𝑇3 – 𝑇2 = 11 – 7 = 4

Substitute 𝑎 = 4 into the formula: 𝑇𝑛 = 𝑎𝑛 + 𝑏 ∴ 𝑇𝑛 = 4𝑛 + 𝑏

Substitute 𝑛 = 1 and 𝑇𝑛 = 3 into the equation, since 𝑇1 = 3: ∴ 3 = 4(1) + 𝑏 ∴ 𝑏 = –1

Substitute 𝑏 = –1 to find the equation: 𝑇𝑛 = 4𝑛 – 1

Then, substitute 105 for 𝑇𝑛 and see if you get an integer answer.

Substitute: 𝑇𝑛 = 105 ∴ 105 = 4𝑛 – 1

Solve for 𝑛:
∴ 𝑛 = 26½

Since 26½ is NOT an integer, it is not possible to have the term 105 in the given sequence.4. Simple Quadratic Sequences

Quadratic sequences are characterised by a second difference that is constant. The general form of the 𝑛th term for a quadratic sequence is:

𝑎𝑛 = 𝑎𝑛² + 𝑏𝑛 + 𝑐

𝑎𝑛 represents the nth term of the sequence. 𝑛 is the position of the term in the sequence. 𝑎, 𝑏 and 𝑐 are constants that need to be determined. Consider the quadratic sequence 3, 8, 15, 24,...

First, identify the first difference: 5, 7, 9,... Then, observe the second difference, which is constant: 2, 2,... To find the formula, we start with the assumption that the sequence is in the form 𝑎𝑛² + 𝑏𝑛 + 𝑐. Using the sequence and the position of the terms, we can set up equations and solve for 𝑎, 𝑏 and 𝑐. For this sequence, the formula turns out to be 𝑛² + 2𝑛, which can be verified by plugging in the values of 𝑛.

Worked Example 8

Find the 𝑛th term of the quadratic sequence: 6, 11, 18, 27,...

Continued Solution to Worked Example 8:

To find the 𝑛th term of the quadratic sequence 6, 11, 18, 27, … we'll follow a step-by-step approach. This involves identifying the pattern, forming equations based on the general form of a quadratic sequence, solving for the constants, and then applying the formula to find any term in the sequence.

Step 1: Identify the Pattern

First, we calculate the first and second differences between the terms to confirm it's a quadratic sequence.

Sequence: 6, 11, 18, 27, … First difference: 11 − 6 = 5, 18 − 11 = 7, 27 − 18 = 9, ... Second difference: 7 − 5 = 2, 9 − 7 = 2, ... The second difference is constant, confirming it's a quadratic sequence.

Step 2: Form Equations

The general form of the 𝑛th term for a quadratic sequence is: 𝑎𝑛 = 𝑎𝑛² + 𝑏𝑛 + 𝑐

To find 𝑎, 𝑏 and 𝑐, we use the first three terms of the sequence and their positions as 𝑛.

For 𝑛 = 1 : 6 = 𝑎(1)² + 𝑏(1) + 𝑐 For 𝑛 = 2 : 11 = 𝑎(2)² + 𝑏(2) + 𝑐 For 𝑛 = 3 : 18= 𝑎(3)² + 𝑏(3) + 𝑐

This gives us three equations:

𝑎 + 𝑏 + 𝑐 = 6 4𝑎 + 2𝑏 + 𝑐 = 11 9𝑎 + 3𝑏 + 𝑐 = 18 Step 3: Solve for Constants

Let's solve these equations for 𝑎, 𝑏 and 𝑐. The solution to the equations is 𝑎 = 1, 𝑏 = 2 and 𝑐 = 3. This means the 𝑛th term formula for the sequence is:

𝑎𝑛 = 𝑛² + 2𝑛 + 3

Step 4: Apply the Formula

Now that we have the formula for the 𝑛th term of the sequence, we can use it to find any term in the sequence. The formula is:

𝑎𝑛 = 𝑛² + 2𝑛 + 3

This formula allows us to calculate the value of any term in the sequence by substituting the term's position for 𝑛. For example, to find the 5th term, substitute 𝑛 = 5:

𝑎5 = 5² + 2(5) + 3 = 25 + 10 + 3 = 38

Thus, the 5th term of the sequence is 38. This method can be applied to find any term in the quadratic sequence 6, 11, 18, 27, … using the derived nth term formula.

5. Simple Cubic Sequences

Cubic sequences have a third difference that is constant, and the general form of the 𝑛th term for a cubic sequence is:

𝑎𝑛 = 𝑎𝑛³ + 𝑏𝑛² + 𝑐𝑛 + 𝑑

𝑎𝑛 represents the nth term of the sequence. 𝑛 is the position of the term in the sequence. 𝑎, 𝑏, 𝑐 and 𝑑 are constants. Consider the cubic sequence 2, 9, 28, 65,...

First, calculate the first difference: 7, 19, 37,... Then, determine the second difference: 12, 18,... Finally, identify the constant third difference: 6,... With the knowledge that the sequence is cubic, we assume the general form 𝑎𝑛 = 𝑎𝑛³ + 𝑏𝑛² + 𝑐𝑛 + 𝑑. By applying the positions of the terms and their values, equations can be formed to solve for 𝑎, 𝑏, 𝑐 and 𝑑. For this sequence, the formula is found to be 𝑛³ + 𝑛, which is validated by inserting the values of 𝑛.

Worked Example 9

Determine the 𝑛th term of the cubic sequence: 1, 8, 27, 64,...

Continued Solution to Worked Example 9:

To determine the 𝑛th term of the cubic sequence 1, 8, 27, 64, … which follows a pattern of cubic numbers, we approach it by confirming its nature and deriving the formula for the 𝑛th term. This sequence represents the cubes of natural numbers, starting from 1³, 2³, 3³, 4³, and so on.

Step 1: Identifying the Pattern

The given sequence is 1, 8, 27, 64, … which corresponds to the cubic numbers of the sequence 1, 2, 3, 4, ... Thus, it's evident that the sequence is cubic due to the nature of the numbers being perfect cubes.

Step 2: Formulate the General Form

For cubic sequences, the general form of the 𝑛th term is:

𝑎𝑛 = 𝑎𝑛³ + 𝑏𝑛² + 𝑐𝑛 + 𝑑

However, since the sequence given is a perfect cube sequence, the formula simplifies to the cube of the position of the term in the sequence, which is:

𝑎𝑛 = 𝑛³

This is because the sequence directly represents the cube of each natural number in order, starting from 1.

Step 3: Verification

To verify, we can apply this formula to the first few terms of the sequence:

For 𝑛 = 1, 𝑎1 = (1)³ = 1 For 𝑛 = 2, 𝑎2 = (2)³ = 8 For 𝑛 = 3, 𝑎3 = (3)³ = 27 For 𝑛 = 4, 𝑎4 = (4)³ = 64 This matches the given sequence exactly, confirming the formula's accuracy.

Step 4: Apply the Formula

The 𝑛th term of the sequence can be found by substituting any natural number for 𝑛 in the formula 𝑎𝑛 = 𝑛³. This allows us to calculate any term in the sequence without having to list all preceding terms.

For instance, to find the 5th term of the sequence:

𝑎5 = 5³ = 125

Thus, the 5th term of the sequence is 125. This approach can be used to find any term in the cubic sequence 1, 8, 27, 64, … by simply cubing the position number 𝑛.

This example demonstrates a straightforward application of recognizing patterns in sequences and formulating a general formula to determine the 𝑛th term, particularly in sequences that follow a specific mathematical operation like cubing.