Answer
1. **(a)** The *nominal* APR is 18% per year (1.5% per month).
The *effective* annual rate is \(\bigl(1.015\bigr)^{12} - 1 \approx 19.55\%\).
2. **(b)** The balance after \(n\) months is
\[
B_{n} = 10{,}000 \;-\; 7{,}000\,(1.015)^{n}.
\]
(Recall \(B_0=3000\).)
3. **(c)** It takes \(\boxed{24\text{ months}}\) to pay off (on the 24th payment).
4. **(d)** The total amount actually paid is about \$3596 if the final payment is *just enough* to clear the balance; or \$3600 if one pays a full \$150 in the last month.
Work Step by Step
Below is a step‐by‐step solution outline for each part (a)–(d). The key points are:
- The card charges 1.5% per month (i.e.\ 0.015 per month), even though it is *nominally* quoted at 18% per year.
- The borrower starts with a \$3000 balance and pays \$150 each month (with *no additional charges*).
- We track the monthly balance using the standard “add interest, then subtract payment” recurrence.
---
## (a) **Annual Percentage Rate (APR)**
Often in textbooks, “APR” can mean simply the *nominal* annual rate, which here is \(12\times 1.5\% = 18\%\). However, many definitions of APR ask for the *effective* annual rate, i.e.\ how much the balance grows in one year if no payments were made.
- The *monthly* growth factor is \(1 + 0.015 = 1.015\).
- Over 12 months, the *effective* annual growth factor is \(\bigl(1.015\bigr)^{12}\).
- Hence the effective annual interest rate is
\[
\bigl(1.015\bigr)^{12} \;-\; 1
\;\approx\; 0.1955
\;=\; 19.55\%.
\]
Depending on your course/text’s definition of APR, the answer is either:
1. **Nominal APR** \(= 18\%\), or
2. **Effective APR** \(\approx 19.55\%\).
---
## (b) **Balance Recurrence and Explicit Formula**
Let \(B_n\) be the balance *after* the \(n\)th monthly payment. We start at
\[
B_0 = 3000 \quad (\text{the initial debt}).
\]
Each month:
1. The old balance \(B_{n-1}\) accrues 1.5% interest: it becomes \(1.015\,B_{n-1}\).
2. The borrower then pays \$150, reducing the balance by 150.
Hence
\[
B_n \;=\; 1.015\,B_{n-1} \;-\; 150,
\quad\text{for } n \ge 1,
\]
with \(B_0 = 3000\).
This is a linear (first‐order) difference equation of the form
\[
B_{n} = r\,B_{n-1} - P,
\quad\text{where } r=1.015 \text{ and } P=150.
\]
### Solving the Recurrence
1. **Homogeneous part**: Solve \(B_{n}^{(h)} = 1.015\,B_{n-1}^{(h)}.\)
A general homogeneous solution is
\[
B_{n}^{(h)} = C \,(1.015)^{n}.
\]
2. **Particular solution**: Guess \(B_{n}^{(p)} = A\) is a constant.
Plugging into \(B_{n} = 1.015\,B_{n-1} - 150\) gives
\[
A \;=\; 1.015\,A \;-\; 150
\quad\Longrightarrow\quad
A - 1.015\,A = -150
\quad\Longrightarrow\quad
-0.015\,A = -150
\quad\Longrightarrow\quad
A = \frac{150}{0.015} \;=\; 10{,}000.
\]
3. **General solution**:
\[
B_{n} = B_{n}^{(h)} + B_{n}^{(p)}
= C \,(1.015)^{n} \;+\; 10{,}000.
\]
4. **Use \(B_{0}=3000\) to find \(C\)**:
\[
B_{0} = 3000
= C \,(1.015)^{0} + 10{,}000
= C + 10{,}000
\quad\Longrightarrow\quad
C = 3000 - 10{,}000 = -7{,}000.
\]
Hence the explicit formula is
\[
\boxed{
B_{n}
= 10{,}000
\;-\; 7{,}000\,\bigl(1.015\bigr)^{n}.
}
\]
---
## (c) **How Long to Pay Off the Debt?**
We want the *smallest* integer \(n\) for which \(B_{n} \le 0\). From the closed‐form solution,
\[
B_{n} \le 0
\quad\Longleftrightarrow\quad
10{,}000 \;-\; 7{,}000\,(1.015)^{n} \;\le\; 0
\quad\Longleftrightarrow\quad
7{,}000\,(1.015)^{n} \;\ge\; 10{,}000
\quad\Longleftrightarrow\quad
(1.015)^{n} \;\ge\; \frac{10{,}000}{7{,}000}
\;=\;\frac{10}{7}.
\]
Taking natural logarithms,
\[
n \;\ge\; \frac{\ln\!\bigl(\tfrac{10}{7}\bigr)}{\ln(1.015)}.
\]
Numerically,
\[
\ln\!\bigl(\tfrac{10}{7}\bigr) \approx 0.356675,\quad
\ln(1.015) \approx 0.014888,\quad
\frac{0.356675}{0.014888} \approx 23.95.
\]
Since \(n\) must be an integer, \(n = 24\) is the *first* time \(B_{n}\) becomes nonpositive.
Therefore, it takes **24 months** to pay off (i.e.\ on the 24th payment, the debt goes to zero or below).
---
## (d) **Total Amount of Money Paid**
In a real‐world setting, one typically makes *equal* payments of \$150 until the final month, when one pays *only* what is left. Let us see what happens in month 24:
- **Balance after 23 payments:**
\[
B_{23} \;=\; 10{,}000 \;-\; 7{,}000\,(1.015)^{23}.
\]
Numerically, \((1.015)^{23}\approx 1.408\), so
\(\;7{,}000\times 1.408 \approx 9856,\)
and thus \(B_{23}\approx 10{,}000 - 9856 = \$144\).
- **Interest added going into month 24:** Multiply \(B_{23}\) by 1.015:
\[
B_{23}\times 1.015 \;\approx\; 144 \times 1.015 \;\approx\; 146.16.
\]
That is how much is owed *just before* the 24th payment.
- **24th payment:** If the person pays exactly \$150, they “overpay” by a few dollars. More typically, one would pay exactly \$146.16 and be done.
Hence there are **23 full payments** of \$150 each, plus a **final (24th) payment** of about \$146. So the *total* paid is approximately
\[
23 \times 150 \;+\; 146
\;=\; 3450 \;+\; 146
\;=\; 3596 \text{ dollars (approximately).}
\]
(If instead the credit‐card agreement requires a *full* \$150 in the last month, then the total outlay would be \$3600, with a small overpayment refunded or simply unneeded. Different banks handle that final payment differently.)
