In Chapter 3, linear functions were applied to examples of linear growth and decay . As seen in the last lesson of this chapter , arithmetic sequences can also be used to model linear growth. Hence, arithmetic sequences can be applied in many areas of life, including simple interest earnings, straight-line depreciation, monthly rental accumulation and many others.

For example, when someone is saving money in equal installments, the cumulative savings at each savings period form an arithmetic sequence. If the driver of a vehicle is travelling down a highway at a constant speed, the amount of petrol left in the tank, measured every minute of the trip, forms another arithmetic sequence. In fact, any time a quantity is changing by equal amounts at set time periods, the process can be considered as being arithmetic and therefore represented by an arithmetic sequence.

Examples

Example 1

A racing car starts the race with 150 litres of fuel. From there, it uses fuel at a rate of 5 litres per minute.

What is the rate of change?

Worked Solution

Create a strategy

The rate of change is the rate at which the fuel increases or decreases every minute.

Apply the idea

Since the fuel is decreasing by 5 litres every minute, we must represent it with negative sign.

\text{Rate of change}=-5 \text{ litres per minute}

Fill in the table of values:

\text{Number of minutes passed ($x$)}	0	5	10	15
\text{Amount of fuel left in tank in litres ($y$)}

Worked Solution

Create a strategy

Find the common difference to find the next terms after 150.

Apply the idea

Recall that every minute, 5 litres are decreasing. So when x=5, or 5 minutes have passed, the amount of fuel in litres decreased is -5\times5=-25.

So the common difference is -25.

When x=0 or when the car starts, the amount of fuel in litres is y=150.

When x=5, the amount of fuel in litres is y=150-25=125.

When x=10, the amount of fuel in litres is y=125-25=100.

When x=15, the amount of fuel in litres is y=100-25=75.

\text{Number of minutes passed (x)}	0	5	10	15
\text{Amount of fuel left in tank in litres (y)}	150	125	100	75

Write an algebraic equation linking the number of minutes passed (x), and the amount of fuel left in the tank (y).

Worked Solution

Create a strategy

Let y be equal to the sum of the first y-value, 150, and the rate of change.

Apply the idea

Since x represents every minute, the rate of change -5 litres per minute can be translated as -5\times x or -5x.

y=150-5x

By rearranging the equation found in part (d), calculate how long it will take for the car to run out of fuel.

Worked Solution

Create a strategy

Substitute y=0 into y=150-5x.

Apply the idea

\displaystyle y	\displaystyle =	\displaystyle 150-5x	Write the equation
\displaystyle 0	\displaystyle =	\displaystyle 150-5x	Susbtitute y=0
\displaystyle 0-150	\displaystyle =	\displaystyle 150-5x-150	Subtract 150 from both sides
\displaystyle -150	\displaystyle =	\displaystyle -5x	Evaluate
\displaystyle \dfrac{-150}{-5}	\displaystyle =	\displaystyle \dfrac{-5x}{-5}	Divide both sides by -5
\displaystyle 30	\displaystyle =	\displaystyle x	Evaluate
\displaystyle x	\displaystyle =	\displaystyle 30 \text{ minutes}	Make x the subject

Example 2

A car bought at the beginning of 2009 is worth \$1500 at the beginning of 2015. The value of the car has depreciated by a constant amount of \$50 each year since it was purchased.

What was the car purchased for in 2009?

Worked Solution

Create a strategy

Multiply the gap year by the yearly depreciation and add the product to the given price.

Apply the idea

The gap year between 2009 and 2015 is 2015-2009=6.

\displaystyle \text{Depreciation amount}	\displaystyle =	\displaystyle 6 \times 50	Multiply the gap year by the depreciation
	\displaystyle =	\displaystyle \$ 300	Evaluate

\displaystyle \text{Purchase price}	\displaystyle =	\displaystyle 1500+300	Add the given price to the depreciation amount
	\displaystyle =	\displaystyle \$1800	Evaluate

Plot the value of the car, V_n, on the graph from 2009 (represented by n=0) to 2015 (represented by n=6.

Worked Solution

Create a strategy

Subtract 50 from each value generated starting from the purchase price.

Apply the idea

For n=0, the value will be the purchase price (1800), so the first point is at (0,1800).

For n=1, the value is 1800-50=1750, so the point is at (1,1750).

For n=2, the value is 1750-50=1700, so the point is at (2,1700).

For n=3, the value is 1700-50=1650, so the point is at (3,1650).

For n=4, the value is 1650-50=1600, so the point is at (4,1600).

For n=5, the value is 1600-50=1550, so the point is at (5,1550).

For n=6, the value is 1550-50=1500, so the point is at (6,1500).

1000

1100

1200

1300

1400

1500

1600

1700

1800

1900

2000

\text{value}

Write an explicit rule for the value of the car after n years. Give the rule in its expanded form.

Worked Solution

Create a strategy

Substitute the values required from part (b) into the explicit formula T_n=a+(n-1)\times d.

Apply the idea

The pronumeral a is the value when n=1, which is 1750 as seen from the graph in part (b).

The common difference d is the amount decreases each year which is -50.

\displaystyle T_n	\displaystyle =	\displaystyle a+(n-1)\times d	Write the formula
\displaystyle T_n	\displaystyle =	\displaystyle 1750+(n-1)\times -50	Substitute a=1750 and d=-50
\displaystyle T_n	\displaystyle =	\displaystyle 1750-50n+50	Expand the brackets
\displaystyle T_n	\displaystyle =	\displaystyle 1800-50n	Collect like terms

Solve for the year n at the end of which the car will be worth half the price it was bought for.

Worked Solution

Create a strategy

Substitute T_n=\dfrac{\text{purchase price}}{2} into T_n=1800-50n.

Apply the idea

\displaystyle T_n	\displaystyle =	\displaystyle \dfrac{\text{purchase price}}{2}	Divide the purchase price by 2
	\displaystyle =	\displaystyle \dfrac{\text{1800}}{2}	Substitute \text{purchase price}=1800
	\displaystyle =	\displaystyle 900	Evaluate

\displaystyle T_n	\displaystyle =	\displaystyle 1800-50n	Write the equation
\displaystyle 900	\displaystyle =	\displaystyle 1800-50n	Substitute T_n=900
\displaystyle 900-1800	\displaystyle =	\displaystyle 1800-50n-1800	Subtract 1800 from both sides
\displaystyle -900	\displaystyle =	\displaystyle -50n	Evaluate
\displaystyle \dfrac{-900}{-50n}	\displaystyle =	\displaystyle \dfrac{-50n}{-50}	Divide both sides by -50
\displaystyle 18	\displaystyle =	\displaystyle n	Evaluate
\displaystyle n	\displaystyle =	\displaystyle 18 \text{ years}	Make n the subject

Idea summary

We can apply the concept of arithmetic sequences to situations that involve linear growth or linear decay.

Hence, a graph of an arithmetic sequence is characterized by a linear growth or linear decay.

Outcomes

U1.AoS2.7

use a given recurrence relation to generate a sequence, deduce the explicit rule, n u from the recursion relation, tabulate, graph and evaluate the sequence

U1.AoS2.8

use a recurrence relation, table or graph to model and analyse practical situations involving discrete linear growth or decay

U1.AoS2.9

demonstrate the use of a recurrence relation to determine the linear depreciating value of an asset after 𝑛 time periods for the initial sequence

U1.AoS2.10

use a rule for the future value of a linear depreciating asset to solve practical problems

5.04 Applications of arithmetic sequences

Ideas

Applications of arithmetic sequences

Examples

Example 1

Example 2

Outcomes

U1.AoS2.7

U1.AoS2.8

U1.AoS2.9

U1.AoS2.10

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