Grade9 Unit 1: Sequences
1. Understanding Sequences
A sequence is an ordered list of numbers. Each number in the list is called a term. Sequences are important in mathematics because they help us identify patterns and make predictions.
Describing Terms in a Sequence
You can use u1 to represent the first term of the sequence, u2 to represent the second term, and so on. The subscripts 1, 2, 3, etc., match the term number. For example, in the sequence 2, 4, 6, 8, 10, …:
- 1st term,
u1 = 2 - 2nd term,
u2 = 4 - 3rd term,
u3 = 6
2. Describing a Sequence Using a Formula
1. Explicit Formula
An explicit formula uses the term’s position number, n, to calculate its value. For example, the formula:
un = 2n + 3 for n ≥ 1tells us that the value of the n-th term (un) is given by 2n + 3 for any value of n greater than or equal to 1. When you are working with sequences, n is always an integer.
Example 1:
A sequence is given by the explicit formula un = 4n - 5 for n ≥ 1.
(a) Find:
(i) the first, second, and fifth terms of the sequence
(ii) the term of the sequence with value 95
(b) Determine whether or not 67 is a term of the sequence.
Solution to Example 1:
(a)
(i) u1 = 4(1) - 5 = -1
u2 = 4(2) - 5 = 3
u5 = 4(5) - 5 = 15
(ii) Solve 4n - 5 = 95:
4n - 5 = 95
4n = 100
n = 25
So, the 25th term is 95.
(b) To determine if 67 is a term:
4n - 5 = 67
4n = 72
n = 18
So, 67 is the 18th term of the sequence.
2. Recursive Formula
A recursive formula gives the relationship between consecutive terms. When you know one term, you can work out the next. For example, the recursive formula:
un+1 = un + 4, u1 = 2 for n ≥ 1Substituting n = 1 gives the 2nd term:
u1 + 1 = u1 + 4 ⇒ u2 = 2 + 4 = 6
Substituting n = 2 gives the 3rd term:
u2 + 1 = u2 + 4 ⇒ u3 = 6 + 4 = 10
Example 2:
A sequence is given by the recursive formula un+1 = un × 2 and u1 = 3.
(a) Find:
(i) the second, third, and fourth terms of the sequence
(ii) the 7th term of the sequence
Solution to Example 2:
(a)
(i) u2 = u1 × 2 = 3 × 2 = 6
u3 = u2 × 2 = 6 × 2 = 12
u4 = u3 × 2 = 12 × 2 = 24
(ii) Using the recursive formula:
u5 = u4 × 2 = 24 × 2 = 48
u6 = u5 × 2 = 48 × 2 = 96
u7 = u6 × 2 = 96 × 2 = 192
So, the 7th term is 192.
3. Linear Sequences
For a linear sequence, the difference between consecutive terms is constant, and the explicit formula is of the form un = a + bn.
A general formula for a sequence is a rule that can be used to generate each term. Usually, the general formula is an explicit formula.
Example 1:
Find a general formula for un, the n-th term of the sequence 5, 10, 15, 20, …
Solution to Example 1:
The common difference d = 10 - 5 = 5.
The first term u1 = 5.
General formula: un = 5 + (n-1)5 = 5n.
Example 2:
The first five terms of a linear sequence are 7, 14, 21, 28, 35.
(a) Find the 50th term of the sequence.
(b) Show that 175 is a member of the sequence and find its term number.
Solution to Example 2:
(a) The common difference d = 14 - 7 = 7.
General formula: un = 7n.
The 50th term: u50 = 7 × 50 = 350.
(b) To check if 175 is a term:
Solve 7n = 175:
n = 25
So, 175 is the 25th term.
4. Recognizing Patterns in Real-Life Contexts
Sequences can be found in various real-life contexts such as predicting earthquakes, financial markets, and many others. Recognizing these patterns can help in making predictions and decisions.
Example 1:
A water tank is initially empty. Water is added at a rate of 5 liters per minute.
(a) Explain why after 3 minutes the tank has 15 liters of water.
(b) Find a general formula for the amount of water in the tank after n minutes.
(c) Determine how many minutes it will take for the tank to have 120 liters of water.
Solution to Example 1:
(a) After 3 minutes:
5 × 3 = 15 liters
(b) General formula:
Wn = 5n, where Wn is the amount of water after n minutes.
(c) To find when the tank has 120 liters:
5n = 120
n = 24
It will take 24 minutes.
Example 2:
A car travels at a constant speed of 60 km/h.
(a) Explain why after 2 hours the car has traveled 120 km.
(b) Find a general formula for the distance traveled after t hours.
(c) Determine how many hours it will take for the car to travel 300 km.
Solution to Example 2:
(a) After 2 hours:
60 × 2 = 120 km
(b) General formula:
Dt = 60t, where Dt is the distance traveled after t hours.
(c) To find when the car travels 300 km:
60t = 300
t = 5
It will take 5 hours.
5. Solving Problems Involving Sequences in Real-Life Contexts
Many real-life problems can be modeled using sequences. For instance, any situation which features a quantity growing at a constant rate can be modeled using a linear function.
Example 1:
A gardener charges a $20 fee to cover transportation expenses, and then $15 per hour for labor while working.
(a) Form an equation for the gardener’s charge C in terms of the time taken, t hours.
(b) Use your equation to determine the cost of a three-hour visit.
(c) Use your equation to determine the cost of a visit lasting 1.5 hours.
(d) Use your equation to determine the length of a visit costing $65.
Solution to Example 1:
(a) Equation:
C = 20 + 15t
(b) Cost for 3 hours:
C = 20 + 15(3) = 65
(c) Cost for 1.5 hours:
C = 20 + 15(1.5) = 42.50
(d) To find the length of a visit costing $65:
20 + 15t = 65
15t = 45
t = 3
So, the visit lasts 3 hours.
Example 2:
A technician charges a $30 fee for the first hour and $20 for each additional hour.
(a) Form an equation for the technician’s charge C in terms of the time taken, t hours.
(b) Use your equation to determine the cost of a five-hour visit.
(c) Use your equation to determine the cost of a visit lasting 2.5 hours.
(d) Use your equation to determine the length of a visit costing $110.
Solution to Example 2:
(a) Equation:
C = 30 + 20(t - 1)
(b) Cost for 5 hours:
C = 30 + 20(5 - 1) = 110
(c) Cost for 2.5 hours:
C = 30 + 20(2.5 - 1) = 60
(d) To find the length of a visit costing $110:
30 + 20(t - 1) = 110
20(t - 1) = 80
t - 1 = 4
t = 5
So, the visit lasts 5 hours.