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Select the correct answer.

Each month, Barry makes three transactions in his checking account:

- He deposits [tex]$\$ 700$[/tex] from his paycheck.
- He withdraws [tex]$\$ 150$[/tex] to buy gas for his car.
- He withdraws [tex]$\$ 400$[/tex] for other expenses.

If his account balance is [tex]$\$ 1,900$[/tex] at the end of the 1st month, which recursive equation models Barry's account balance at the end of month [tex]$n$[/tex]?

A. [tex]f(1) = 1,900[/tex]
[tex]f(n) = f(n-1) + 150, \text{ for } n \geq 2[/tex]

B. [tex]f(1) = 1,900[/tex]
[tex]f(n) = f(n-1) + 700, \text{ for } n \geq 2[/tex]

C. [tex]f(1) = 1,900[/tex]
[tex]f(n) = f(n-1) - 150, \text{ for } n \geq 2[/tex]

D. [tex]f(1) = 1,900[/tex]
[tex]f(n) = 150 \cdot f(n-1), \text{ for } n \geq 2[/tex]

Answer :

To determine which recursive equation models Barry's account balance each month, let's take a closer look at the transactions and how they affect his balance.

Barry's monthly transactions are as follows:
- He deposits [tex]$700 into his account.
- He withdraws $[/tex]150 for gas.
- He withdraws [tex]$400 for other expenses.

Let's calculate the net change in Barry's account balance each month:

1. Deposits: Barry receives $[/tex]700.
2. Withdrawals: Barry loses [tex]$150 for gas and $[/tex]400 for other expenses, totaling [tex]$550 in withdrawals.
3. Net Change: The net change in his account balance per month is $[/tex]700 (deposit) minus [tex]$550 (total withdrawals), which equals a net increase of $[/tex]150.

Given this net change, we can now determine the recursive equation:

- His initial balance at the end of the 1st month is [tex]$1,900, as given.
- Each subsequent month's balance is the previous month's balance plus the net change of $[/tex]150.

Thus, the recursive equation that models Barry's account balance at the end of month [tex]\( n \)[/tex] is:

- [tex]\( f(1) = 1,900 \)[/tex]
- [tex]\( f(n) = f(n-1) + 150, \text{ for } n \geq 2 \)[/tex]

Looking at the answer choices, the correct option that matches this model is:

A.
[tex]\( f(1) = 1,900 \)[/tex]
[tex]\( f(n) = f(n-1) + 150, \text{ for } n \geq 2 \)[/tex]

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