Applications of arithmetic sequences
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 instalments, 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.
Worked example
Example 1
Tabitha starts with $\$200$$200 in her piggy bank, the following week she adds $\$25$$25 and then continues to add $\$25$$25 at the start of each successive week. Find a rule to describe $B_n$Bn the balance of her savings at the start of each week and find when her savings will reach $\$450$$450.
Think: The sequence of savings generated is $\$200,\$225,\$250,\$275...$$200,$225,$250,$275... This is arithmetic, so write down the starting value $a$a and common difference and use $u_n=a+\left(n-1\right)d$un=a+(n−1)d to find a general rule.
Do: $a=200$a=200 and $d=25$d=25 and so our general rule is: $B_n=200+25(n-1)$Bn=200+25(n−1) or equivalently $B_n=175+25n$Bn=175+25n.
To find when the savings reach $\$450$$450, we substitute this into our general rule and solve for $n$n:
| $450$450 | $=$= | $175+25n$175+25n |
| $\therefore25n$∴25n | $=$= | $275$275 |
| $n$n | $=$= | $11$11 |
Hence, at the start of the $11$11th week her savings will have grown to $\$450$$450.
Practice questions
QUESTION 1
A racing car starts the race with $150$150 litres of fuel. From there, it uses fuel at a rate of $5$5 litres per minute.
What is the rate of change?
Fill in the table of values:
Number of minutes passed ($x$x) $0$0 $5$5 $10$10 $15$15 Amount of fuel left in tank in litres ($y$y) $150$150 $\editable{}$ $\editable{}$ $\editable{}$ Write an algebraic equation linking the number of minutes passed ($x$x) and the amount of fuel left in the tank ($y$y).
By rearranging the equation found in part (d), calculate how long it will take for the car to run out of fuel.
QUESTION 2
A car bought at the beginning of 2009 is worth $\$1500$$1500 at the beginning of 2015. The value of the car has depreciated by a constant amount of $\$50$$50 each year since it was purchased.
What was the car purchased for in 2009?
Plot the value of the car, $V_n$Vn, on the graph from 2009 (represented by $n=0$n=0) to 2015 (represented by $n=6$n=6).
Loading Graph...Write an explicit rule for the value of the car after $n$n years.
Give the rule in its expanded form.
Solve for the year $n$n at the end of which the car will be worth half the price it was bought for.