Suppose your business has a special checking account used just for paying the phone bill. The balance is $740.00 this month. Next month the balance will be $707.60, after that it will be $675.20, and on the third month the balance will be $642.80. Write an explicit formula 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? Question 9 options: A(n) = 740.00 + (n - 1)(-32.40); 24 months A(n) = 740.00 - 32.40n; 22 months A(n) = 740.00 + (n - 1)(-32.40); 23 months A(n) = 740.00 - 32.40n; 23 months

An arithmetic sequence is defined by a starting value and a common difference, and once you have those values, you can write:
[tex]A_n = s - d(n-1)[/tex], with [tex]s[/tex] being the starting value and [tex]d[/tex] being the common difference.

We know that the starting value is [tex]$740[/tex], since that's the balance this month.  The common difference is the difference between two consecutive terms, which is

[tex]$707.60-$740 = -$32.40[/tex]

Therefore, the arithmetic sequence is [tex]A_n = $740 - $32.40 \cdot (n-1)[/tex].

This corresponds to the first and third options.  However, the question of how many months you can pay the bills remains.

We can no longer pay our bills when the balance becomes zero or less, so we solve for the greatest value of [tex]n[/tex] satisfying

[tex]$740 - $32.40 \cdot (n-1) \geq 0[/tex]. 

Expanding, this becomes

[tex]$772.40 - $32.40 \cdot n \geq 0[/tex],

which is equivalent to

[tex]$772.40 \geq $32.40 \cdot n[/tex].  Dividing by [tex]$32.40[/tex] gives
[tex]23.83 \geq n[/tex] (the [tex]23.83[/tex] is an approximation, but it's close enough for our purposes). 
Since [tex]n[/tex] has to be an integer, the maximum value of [tex]n[/tex] is [tex]\boxed{23}[/tex].  Thus, the third option is correct.  (
A(n) = 740.00 + (n - 1)(-32.40); 23 months)