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Linear Equations and Arithmetic Sequences . Lesson 3.1. You can solve many rate problems by using recursion.
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Linear Equations and Arithmetic Sequences Lesson 3.1
You can solve many rate problems by using recursion. Matias wants to call his aunt in Chile on her birthday. He learned that placing the call costs $2.27 and that each minute he talks costs $1.37. How much would it cost to talk for 30 minutes? You can calculate the cost of Matias’s phone call with the recursive formula
To find the cost of a 30-minute phone call, calculate the first 30 terms, as shown in the calculator screen.
As you learned in algebra, you or Matias can also find the cost of a 30-minute call by using the linear equation y = 2.27 +1.37x where x is the length of the phone call in minutes and y is the cost in dollars. If the phone company always rounds up the length of the call to the nearest whole minute, then the costs become a sequence of discrete points, and you can write the relationship as an explicit formula, un = 2.27+ 1.37n, where n is the length of the call in whole minutes and un is the cost in dollars.
Example A • Consider the recursively defined arithmetic sequence • Find an explicit formula for the sequence. Notice you start with 2 and keep adding a common difference or rate of change is 6.
In general, when you write the formula for a sequence, you use n to represent the number of the term and un to represent the term itself.
Example A • Consider the recursively defined arithmetic sequence • Use the explicit formula to find u22 .
Example A • Consider the recursively defined arithmetic sequence • Find the value of n so that un = 86.
You graphed sequences of points (n, un ) in Chapter 1. The term number n is a whole number: 0, 1, 2, 3, . . . . So, using different values for n will produce a set of discrete points. The points on this graph show the arithmetic sequence from Example A.
When n increases by 1, un increases by 6, the common difference. The change in the y-value is 6 that corresponds to a change of 1 in the x-value. So the points representing the sequence lie on a line with a slope of 6. In general, the common difference, or rate of change, between consecutive terms of an arithmetic sequence is the slope of the line through those points.
The pair (0, 2) names the starting value 2, which is the y-intercept. Using the intercept form of a linear equation, you can now write an equation of the line through the points of the sequence as y = 2+6x, or y=6x+ 2.
Match Point • Below are three recursive formulas, three graphs, and three linear equations.
Match the recursive formulas, graphs, and linear equations that go together. (Not all of the appropriate matches are listed. If the recursive rule, graph, or equation is missing, you will need to create it.)
Write a brief statement relating the starting value and common difference of an arithmetic sequence to the corresponding equation y=a+bx.
Are points (n, un) of an arithmetic sequence always collinear? Write a brief statement supporting your answer.
Example B • Retta typically spends $2 a day on lunch. She notices that she has $17 left after today’s lunch. She thinks of this sequence to model her daily cash balance. • Find the explicit formula that represents her daily cash balance and an equation of the line through the points of this sequence. Each term is 2 less than the previous term, so the common difference of the arithmetic sequence and the slope of the line are both -2. The term t1 is 17, so the previous term, t0, or the y-intercept, is 19. y = 19 - 2x.
Example B • Retta typically spends $2 a day on lunch. She notices that she has $17 left after today’s lunch. She thinks of this sequence to model her daily cash balance. • How useful is this formula for predicting how much money Retta will have each day? We don’t know whether Retta has any other expenses she might encounter. The formula could be valid for eight more days, until she has $1 left (on t9), as long as she gets no more money and spends only $2 per day.
