College

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]m[/tex]?

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

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

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

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

Answer :

To solve this problem, we're looking for a recursive equation that models Barry's account balance at the end of month [tex]\( m \)[/tex].

1. Understand the Transactions:
- Barry deposits \[tex]$700 each month.
- Barry withdraws \$[/tex]150 for gas.
- Barry also withdraws \[tex]$400 for other expenses.

2. Calculate the Net Monthly Change:
To find out how Barry's account balance changes each month, we calculate the total withdrawals and subtract it from the deposit:
\[
\text{Net monthly change} = \$[/tex]700 - (\[tex]$150 + \$[/tex]400) = \[tex]$700 - \$[/tex]550 = \[tex]$150
\]
This means each month, Barry adds \$[/tex]150 to his account after all transactions.

3. Define the Recursive Equation:
We are given that Barry's account balance at the end of the first month is \[tex]$1,900. So we define:
\[
f(1) = 1900
\]
For each subsequent month \( n \geq 2 \), the account balance is the previous month’s balance plus the net change of \$[/tex]150:
[tex]\[
f(n) = f(n-1) + 150, \text{ for } n \geq 2
\][/tex]

4. Select the Correct Answer:
Looking at the available choices in the problem, the correct recursive equation that matches our findings is:
[tex]\[
\text{Choice D: } f(1) = 1,900; \, f(n) = f(n-1) + 150, \text{ for } n \geq 2
\][/tex]

Thus, the correct answer is D.