Lesson Notes By Weeks and Term v5 - Grade 10
Number patterns – Week 6 focus
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Number patterns – Week 6 focus
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Subject: Mathematics
Class: Grade 10
Term: 1st Term
Week: 6
Theme: General lesson support
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Number patterns are all around us, from the arrangement of tiles on a roof in Soweto to the growth of small businesses in townships and the interest accumulated on savings accounts. Understanding number patterns allows us to predict future values, analyze trends, and solve real-world problems. This week, we'll focus on identifying and describing arithmetic and quadratic number patterns, finding their general rules (or nth term), and using these rules to solve problems. This is a crucial skill for further studies in mathematics and related fields like finance, engineering, and computer science.
Arithmetic Sequences An arithmetic sequence (or arithmetic progression) is a sequence where the difference between consecutive terms is constant. This constant difference is called the common difference (d).
General Form: a, a + d, a + 2d, a + 3d, ...
General Term: T n = a + (n - 1)d Where: T n is the nth term of the sequence. a is the first term of the sequence. n is the term number (position of the term in the sequence). d is the common difference.
Example 1: Consider the sequence: 2, 5, 8, 11, ... Is it arithmetic? Yes, because the difference between consecutive terms is constant (5 - 2 = 3, 8 - 5 = 3, 11 - 8 = 3).
Therefore, d =
3. The first term, a =
2. The general term: T n = 2 + (n - 1)3 = 2 + 3n - 3 = 3n - 1 To find the 10th term: T 10 = 3(10) - 1 = 30 - 1 = 29 Example 2: The monthly repayments on a loan form an arithmetic sequence. If the first repayment is R1500 and each subsequent repayment is R50 less than the previous one, find the 12th repayment. a = 1500 (first term) d = -50 (common difference, since repayments are decreasing) n = 12 (we want the 12th repayment) T n = a + (n - 1)d T 12 = 1500 + (12 - 1)(-50) = 1500 + (11)(-50) = 1500 - 550 = R950 Quadratic Sequences A quadratic sequence is a sequence where the second difference between consecutive terms is constant. This means the differences between consecutive terms (the first differences) form an arithmetic sequence.
General Form: T n = an 2 + bn + c Where a, b, and c are constants. Finding the General Term (T n = an 2 + bn + c): Calculate the first differences: Subtract each term from the next term.
Calculate the second differences: Subtract each first difference from the next first difference. These 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 (the first term of the original sequence) Solve these equations simultaneously to find the values of a, b, and c. Substitute the values of a, b, and c into the general form T n = an 2 + bn + c.
Example 3: Consider the sequence: 2, 7, 14, 23, ...
First Differences: 7 - 2 = 5; 14 - 7 = 7; 23 - 14 = 9 Second Differences: 7 - 5 = 2; 9 - 7 = 2 (Constant!)
Equations: 2a = 2 => a = 1 3a + b = 5 => 3(1) + b = 5 => b = 2 a + b + c = 2 => 1 + 2 + c = 2 => c = -1 General Term: T n = (1)n 2 + (2)n + (-1) = n 2 + 2n - 1 To find the 6th term: T 6 = (6) 2 + 2(6) - 1 = 36 + 12 - 1 = 47 Example 4: Find the general term of the quadratic sequence: 0, 3, 8, 15, ...
First differences: 3, 5, 7 Second differences: 2, 2 Equations: 2a = 2 => a = 1 3a + b = 3 => 3(1) + b = 3 => b = 0 a + b + c = 0 => 1 + 0 + c = 0 => c = -1 General Term: T n = n 2 - 1 Guided Practice (With Solutions)
Question 1: Find the 15th term of the arithmetic sequence: 1, 5, 9, 13, ...
Solution: a = 1 d = 5 - 1 = 4 n = 15 T n = a + (n - 1)d T 15 = 1 + (15 - 1)4 = 1 + (14)4 = 1 + 56 = 57
Commentary: This is a straightforward application of the arithmetic sequence formula. Make sure to correctly identify 'a' and 'd'.
Question 2: The following sequence is quadratic: 3, 6, 11, 18, ... Determine the general term.
Solution: First differences: 3, 5, 7 Second differences: 2, 2 Equations: 2a = 2 => a = 1 3a + b = 3 => 3(1) + b = 3 => b = 0 a + b + c = 3 => 1 + 0 + c = 3 => c = 2 General Term: T n = n 2 + 2
Commentary: Remember the crucial step of finding the first and second differences. Setting up the equations correctly is key to solving for a, b, and c.
Question 3: In an arithmetic sequence, the 3rd term is 7 and the 7th term is
1
5. Find the first term (a) and the common difference (d).
Solution: T 3 = a + 2d = 7 (Equation 1) T 7 = a + 6d = 15 (Equation 2)
Subtract Equation 1 from Equation 2: (a + 6d) - (a + 2d) = 15 - 7 => 4d = 8 => d = 2 Substitute d = 2 into Equation 1: a + 2(2) = 7 => a + 4 = 7 => a = 3
Commentary: This problem requires solving simultaneous equations. Practice setting up these equations correctly based on the given information. Independent Practice (Questions Only) Find the 20th term of the arithmetic sequence: -5, -1, 3, 7, ... Determine the general term (T n ) of the arithmetic sequence: 10, 7, 4, 1, ... The general term of a sequence is T n = 2n 2 - n +
3. Find the first three terms of the sequence. The first three terms of a quadratic sequence are 1, 6, and
1
3. Determine the general term. Which term in the arithmetic sequence 4, 7, 10, ... is equal to 49? The 5th term of an arithmetic sequence is 22, and the 12th term is
5
0. Determine the first term and the common difference. Find the next two terms of the quadratic sequence 2, 5, 10, 17, ... The sequence 5; x; 17; ... is arithmetic. Determine the value of x.
Consider the sequence: 3, 7, 13, 21, 31, …. a) Determine the general term of the sequence. b) Calculate the 20th term of the sequence.
Given the quadratic sequence: -1; 5; 15; 29; … a) Determine the general term T n of the sequence. b) Calculate the value of the 25 th term.