Lesson Notes By Weeks and Term v5 - Grade 11

Lesson Video

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Performance objectives

• Identify a quadratic sequence.
• Determine the general term (Tn) of a quadratic sequence.
• Calculate specific terms in a sequence.
• Solve problems involving quadratic sequences.
• Apply linear equations to find a, b, and c.

Lesson summary

Number patterns are fundamental to mathematics and play a significant role in understanding the world around us. From predicting stock market trends (which affects many South Africans' investments) to understanding population growth, number patterns provide a framework for making informed decisions. Specifically, in Week 6, we will delve into quadratic sequences, a type of number pattern where the second difference between consecutive terms is constant. These patterns are more complex than arithmetic or geometric sequences and require a deeper understanding of algebraic manipulation. Think about the seating arrangements in a stadium - often the rows increase quadratically!

Lesson notes

A quadratic sequence is a sequence of numbers where the second difference between consecutive terms is constant. This means that if you subtract consecutive terms, and then subtract consecutive differences, you'll arrive at a constant value. The general term of a quadratic sequence can be expressed as: T n = an 2 + bn + c where a, b, and c are constants, and n represents the position of the term in the sequence (n = 1, 2, 3, ...). It's critical to understand that a cannot be zero; otherwise, the sequence would become linear. Finding the General Term (T n ) To find the general term, we need to determine the values of a, b, and c. We can do this by using the following relationships derived from the first three terms of the sequence: First Difference: Let the first three terms of the sequence be T 1 , T 2 , and T 3 .

Calculate the first differences: d 1 = T 2 - T 1 d 2 = T 3 - T 2 Second Difference: Calculate the second difference: D = d 2 - d 1 Relationships: The constants a, b, and c are related to the first term (T 1 ), the first difference (d 1 ) and the second difference (D) as follows: 2a = D 3a + b = d 1 a + b + c = T 1 We now have a system of three linear equations with three unknowns (a, b, and c). We can solve this system to find the values of a, b, and c, and hence the general term T n .

Worked example

Example 1: Find the general term of the quadratic sequence: 5, 12, 23, 38, ...

First Differences:

d 1 = 12 - 5 = 7

d 2 = 23 - 12 = 11

d 3 = 38 - 23 = 15

Second Difference:

D = 11 - 7 = 4

(Also, 15-11 = 4, confirming its quadratic)

Relationships:

2a = 4 => a = 2

3a + b = 7 => 3(2) + b = 7 => 6 + b = 7 => b = 1

a + b + c = 5 => 2 + 1 + c = 5 => 3 + c = 5 => c = 2

General Term:

T n = 2n 2 + n + 2

Example 2: A construction company is stacking bricks in a triangular pattern. The number of bricks in each row follows the sequence 1, 3, 6, 10, ... Find the formula for the number of bricks in the nth row.

First Differences:

d 1 = 3 - 1 = 2

d 2 = 6 - 3 = 3

d 3 = 10 - 6 = 4

Second Difference:

D = 3 - 2 = 1

Relationships:

2a = 1 => a = 1/2

3a + b = 2 => 3(1/2) + b = 2 => 3/2 + b = 2 => b = 1/2

a + b + c = 1 => 1/2 + 1/2 + c = 1 => 1 + c = 1 => c = 0

General Term:

T n = (1/2)n 2 + (1/2)n + 0

T n = (n 2 + n)/2

T n = n(n+1)/2

Example 3: Given the sequence -1, 6, 17, 32,..., determine the value of T 10 .

First Differences:

d 1 = 6 - (-1) = 7

d 2 = 17 - 6 = 11

d 3 = 32 - 17 = 15

Second Difference:

D = 11 - 7 = 4

Relationships:

2a = 4 => a = 2

3a + b = 7 => 3(2) + b = 7 => 6 + b = 7 => b = 1

a + b + c = -1 => 2 + 1 + c = -1 => 3 + c = -1 => c = -4

General Term:

T n = 2n 2 + n - 4

Calculate T 10 :

T 10 = 2(10) 2 + 10 - 4

T 10 = 2(100) + 10 - 4

T 10 = 200 + 10 - 4

T 10 = 206

Guided Practice (With Solutions)

Question 1: Determine whether the following sequence is quadratic: 2, 6, 12, 20, ...