Lesson Notes By Weeks and Term v5 - Grade 12
Patterns, sequences and series – Week 1 focus
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Patterns, sequences and series – Week 1 focus
Download the Lessonotes Mobile South Africa app for faster lesson access on Android and iPhone.
Subject: Mathematics
Class: Grade 12
Term: 1st Term
Week: 1
Theme: General lesson support
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Patterns, sequences, and series are fundamental building blocks in mathematics, enabling us to model and understand numerous phenomena in the world around us. In the South African context, understanding these concepts is crucial, from analyzing population growth trends to calculating compound interest on investments or even predicting electricity usage patterns. The ability to identify patterns and make predictions based on them is a powerful tool for problem-solving and critical thinking in various aspects of life. This week, we will focus specifically on arithmetic and quadratic sequences and series.
2. 1.
Sequences: A sequence is an ordered list of numbers called terms. Each term is denoted by T n , where n represents the position of the term in the sequence (e.g., T 1 is the first term, T 2 is the second term, and so on). 2.
2. Arithmetic Sequences: An arithmetic sequence is a sequence where the difference between consecutive terms is constant. This constant difference is called the common difference, denoted by d.
General Term (T n ): T n = a + (n - 1)d Where: a is the first term (T 1 *) n is the term number d is the common difference
Example: Consider the sequence: 2, 5, 8, 11, 14… Here, a = 2 and d = 5 - 2 =
3. The general term is T n = 2 + (n - 1)3 = 3n -
1. To find the 10th term, T 10 = 3(10) - 1 = 29. 2.
3. Arithmetic Series: An arithmetic series is the sum of the terms of an arithmetic sequence.
Sum of the first n terms (S n ): S n = n/2 [2a + (n - 1)d] or S n = n/2 [a + L], where L is the last term.
Example: Consider the arithmetic sequence: 2, 5, 8, 11, 14… To find the sum of the first 5 terms, we can use either formula.
Using the first formula: S 5 = 5/2 [2(2) + (5 - 1)3] = 5/2 [4 + 12] = 5/2 [16] =
4
0. Using the second formula: First, find the 5th term (already given as 14). Then, S 5 = 5/2 [2 + 14] = 5/2 [16] = 40. 2.
4. Quadratic Sequences: A quadratic sequence is a sequence where the second difference between consecutive terms is constant.
General Term (T n ): T n = an 2 + bn + c Where a, b, and c are constants that need to be determined. Method to find a, b, and c: Find the first difference: Subtract each term from the term that follows it.
Find the second difference: Subtract each first difference from the first difference that follows it. The second difference should be constant.
Use the following relationships: 2a = Second Difference 3a + b = First Difference between T 1 and T 2 * a + b + c = T 1 *
Example: Consider the sequence: 1, 3, 7, 13, 21… First Difference: 2, 4, 6, 8… Second Difference: 2, 2, 2… (Constant) Solve for a, b, and c: 2a = 2 => a = 1 3a + b = 2 => 3(1) + b = 2 => b = -1 a + b + c = 1 => 1 + (-1) + c = 1 => c = 1 Therefore, the general term is T n = n 2 - n +
1. To find the 6th term, T 6 = (6) 2 - 6 + 1 = 36 - 6 + 1 = 31. 2.
5. Relationship between Arithmetic and Quadratic Sequences The first difference of a quadratic sequence results in an arithmetic sequence. 2.6 Worked Examples Example 1: Given the arithmetic sequence 4; 7; 10;... a) Determine the general term T n . b) Calculate T 20 . c) Which term is equal to 85?
Solution: a) a=4; d = 7-4 = 3 T n = a + (n-1)d T n = 4 + (n-1)3 T n = 4 + 3n -3 T n = 3n + 1 b) T 20 * = 3(20) + 1 T 20 = 60 + 1 T 20 = 61 c) 85 = 3n + 1 84 = 3n n = 28 Therefore, the 28 th term is equal to
8
5. Example 2: The following sequence is quadratic: 0; 3; 8; 15; ... a) Determine the general term T n . b) Calculate T 15 .
Solution: a) 1st Difference: 3; 5; 7; 2nd Difference: 2; 2; 2a = 2 a = 1 3a + b = 3 3(1) + b = 3 b = 0 a + b + c = 0 1 + 0 + c = 0 c = -1 Therefore, T n = n 2 - 1 b) T 15 = 15 2 * - 1 T 15 = 225 - 1 T 15 = 224 Guided Practice (With Solutions)
Question 1: Determine the 15th term (T 15 ) and the sum of the first 15 terms (S 15 ) of the arithmetic sequence: 1, 5, 9, 13… Solution: Identify a and d: a = 1, d = 5 - 1 =
4. Find T 15 : T 15 = 1 + (15 - 1)4 = 1 + 14(4) = 1 + 56 =
5
7. Find S 15 : S 15 = 15/2 [2(1) + (15 - 1)4] = 15/2 [2 + 56] = 15/2 [58] = 15(29) =
4
3
5. Question 2: The general term of a sequence is given by T n = 2n +
3. Determine whether the sequence is arithmetic, and if so, find the common difference.
Solution: Find the first few terms: T 1 = 2(1) + 3 = 5, T 2 = 2(2) + 3 = 7, T 3 = 2(3) + 3 =
9. Check for a common difference: T 2 - T 1 = 7 - 5 = 2, T 3 - T 2 = 9 - 7 =
2. Since the difference is constant (2), the sequence is arithmetic with a common difference of d =
2. Question 3: The following sequence is quadratic: 2; 6; 12; 20;...Determine the general term (T n ) of this sequence.
Solution: First Difference: 4; 6; 8; Second Difference: 2; 2; 2a = 2 => a = 1 3a + b = 4 => 3(1) + b = 4 => b = 1 a + b + c = 2 => 1 + 1 + c = 2 => c = 0 Therefore, T n = n 2 + n Question 4: Determine S n for the following sequence: 5 + 10 + 15 +...to 12 terms.
Solution: This is an arithmetic series. a = 5; d = 5; n = 12 S n = n/2[2a + (n-1)d] S 12 = 12/2[2(5) + (12 - 1)5] S 12 = 6[10 + (11)5] S 12 = 6[10 + 55] S 12 = 6[65] S 12 = 390 Independent Practice (Questions Only) Determine the general term (T n ) for the arithmetic sequence: 7, 11, 15, 19… Calculate the 25th term (T 25 ) of the arithmetic sequence defined by T n = 5n -
3. Find the sum of the first 20 terms (S 20 ) of the arithmetic series: 3 + 8 + 13 + 18… Given an arithmetic sequence where T 3 = 8 and T 7 = 20, find the first term (a) and the common difference (d). Then, find T 12 .
The following sequence is quadratic: 2; 8; 16; 26;... Determine T n of this sequence.