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) + 700 - 150 - 400[/tex], for [tex]n \geq 2[/tex]

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

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

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

Barry's account balance at the end of the first month is given as

[tex]$$
f(1) = 1900.
$$[/tex]

Each month, he makes three transactions:

1. He deposits \[tex]$700.
2. He withdraws \$[/tex]150 for gas.
3. He withdraws \[tex]$400 for other expenses.

The net change in his account each month is

$[/tex][tex]$
700 - 150 - 400 = 150.
$[/tex][tex]$

This means that starting from month 2, his balance increases by \$[/tex]150 each month. Therefore, the recursive equation that models his account balance is

[tex]$$
f(1)=1900, \quad f(n)=f(n-1)+150 \text{ for } n\geq 2.
$$[/tex]

This corresponds to option D.