Let
A(n) be the representing the amount of money left on the card. At the beginning, the cafeteria card's value is
$50. This means that our first term,
A(1), is
50. After the first purchase on Monday, its value is
$46.75, and after the next day its value is
$43.50.
50 ⟶-3.25 46.75 ⟶-3.25 43.5 ⟶-3.25 …
As we can see, the
d is
-3.25.
Writing an Explicit Formula
Let's recall the form of an of an arithmetic sequence.
A(n)=A(1)+(n−1)d
In the above formula,
A(1) is the initial value and
d is the common difference. We will get our formula by substituting
A(1)=50 and
d=-3.25.
A(n)=A(1)+(n−1)d
A(n)=50+(n−1)(-3.25)
A(n)=50+(-3.25)(n−1)
A(n)=50−3.25(n−1)
This formula gives us the terms of the arithmetic sequence formed by the amount of money left on the card.
Finding the Value of the Card
Let's review our sequence again.
50, 46.75, 43.5, …
Notice that the
2nd term of the sequence is the amount of money left on the card after buying
1 lunch, the
3rd term is the amount of money left on the card after buying
2 lunches, and so on. Therefore, the
13th term of the sequence will give us the amount of money left on the card after buying
12 lunches.
A(n)=50−3.25(n−1)
A(13)=50−3.25(13−1)
A(13)=50−3.25(12)
A(13)=50−39
A(13)=11
This means that after buying
12 lunches, the card's value will be
$11.