Suppose your business has a special checking account used just for paying the phone bill. The balance is $1040.00 this month. Next month the balance will be $995.80, after that it will be $951.60, and on the third month the balance will be $907.40. Write a rule to represent the balance in the account as an arithmetic sequence. How many months can you pay your phone bill without depositing any more money in the account?
Q1
We can get a table :
Balance 1040 995.8 951.6 907.4
month(m) 0 1 2 3
We get the function:
Tm = 1040-44.2m
Q2
1040 > 44.2m
23.5 > m
It can pay for 23 months without deposting any more money.
To write a rule representing the balance in the account as an arithmetic sequence, we need to determine the common difference (d) between consecutive terms.
The first term, a₁, is $1040.00, and the second term, a₂, is $995.80. To find the common difference, we subtract the first term from the second term:
d = a₂ - a₁
= $995.80 - $1040.00
= -$44.20
Therefore, the common difference, d, is -$44.20.
Now that we have the common difference, we can write the rule for the arithmetic sequence:
aₙ = a₁ + (n - 1) * d
where aₙ represents the balance in the account after n months and n is the number of months.
Using the given information, a₁ = $1040.00, and the common difference, d = -$44.20, we can substitute these values into the rule:
aₙ = 1040 + (n - 1) * (-44.20)
= 1040 - 44.20n + 44.20
Simplifying the expression, we get:
aₙ = 1084.20 - 44.20n
Now, to find how many months you can pay your phone bill without depositing any more money in the account, we need to determine when the balance, aₙ, becomes zero.
Setting aₙ = 0, we have:
0 = 1084.20 - 44.20n
Rearranging the equation and solving for n, we get:
44.20n = 1084.20
n = 1084.20 / 44.20
≈ 24.53
Since we cannot have a fractional number of months, we round down to the nearest whole number.
Therefore, you can pay your phone bill without depositing any more money in the account for 24 months.