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Answer :
We start with the annuity formula for the future value of a series of monthly payments:
[tex]$$
A = M \cdot \frac{\left(1 + \frac{r}{n}\right)^{n t} - 1}{\frac{r}{n}},
$$[/tex]
where
- [tex]$A$[/tex] is the total accumulated amount,
- [tex]$M$[/tex] is the monthly payment,
- [tex]$r$[/tex] is the annual interest rate (in decimal form),
- [tex]$n$[/tex] is the number of compounding periods per year, and
- [tex]$t$[/tex] is the time in years.
Given:
[tex]$$
A = 7000, \quad r = 0.054, \quad n = 12, \quad t = 6.
$$[/tex]
Step 1. Calculate the monthly interest rate:
[tex]$$
\frac{r}{n} = \frac{0.054}{12} = 0.0045.
$$[/tex]
Step 2. Determine the total number of compounding periods:
[tex]$$
n t = 12 \times 6 = 72.
$$[/tex]
Step 3. Compute the growth factor in the formula:
[tex]$$
\left(1 + \frac{r}{n}\right)^{n t} - 1 = \left(1 + 0.0045\right)^{72} - 1.
$$[/tex]
Evaluating this expression, we obtain approximately:
[tex]$$
\left(1.0045\right)^{72} - 1 \approx 0.3816427360289367.
$$[/tex]
Step 4. Substitute these values into the annuity formula and solve for the monthly payment [tex]$M$[/tex]. The formula rearranged for [tex]$M$[/tex] is:
[tex]$$
M = \frac{A\cdot \left(\frac{r}{n}\right)}{\left(1 + \frac{r}{n}\right)^{n t} - 1}.
$$[/tex]
Substitute the known values:
[tex]$$
M = \frac{7000 \times 0.0045}{0.3816427360289367}.
$$[/tex]
Step 5. Calculate [tex]$M$[/tex]:
[tex]$$
M \approx \frac{31.5}{0.3816427360289367} \approx 82.5379262494639.
$$[/tex]
Thus, the monthly payment is approximately [tex]$82.54$[/tex].
[tex]$$
A = M \cdot \frac{\left(1 + \frac{r}{n}\right)^{n t} - 1}{\frac{r}{n}},
$$[/tex]
where
- [tex]$A$[/tex] is the total accumulated amount,
- [tex]$M$[/tex] is the monthly payment,
- [tex]$r$[/tex] is the annual interest rate (in decimal form),
- [tex]$n$[/tex] is the number of compounding periods per year, and
- [tex]$t$[/tex] is the time in years.
Given:
[tex]$$
A = 7000, \quad r = 0.054, \quad n = 12, \quad t = 6.
$$[/tex]
Step 1. Calculate the monthly interest rate:
[tex]$$
\frac{r}{n} = \frac{0.054}{12} = 0.0045.
$$[/tex]
Step 2. Determine the total number of compounding periods:
[tex]$$
n t = 12 \times 6 = 72.
$$[/tex]
Step 3. Compute the growth factor in the formula:
[tex]$$
\left(1 + \frac{r}{n}\right)^{n t} - 1 = \left(1 + 0.0045\right)^{72} - 1.
$$[/tex]
Evaluating this expression, we obtain approximately:
[tex]$$
\left(1.0045\right)^{72} - 1 \approx 0.3816427360289367.
$$[/tex]
Step 4. Substitute these values into the annuity formula and solve for the monthly payment [tex]$M$[/tex]. The formula rearranged for [tex]$M$[/tex] is:
[tex]$$
M = \frac{A\cdot \left(\frac{r}{n}\right)}{\left(1 + \frac{r}{n}\right)^{n t} - 1}.
$$[/tex]
Substitute the known values:
[tex]$$
M = \frac{7000 \times 0.0045}{0.3816427360289367}.
$$[/tex]
Step 5. Calculate [tex]$M$[/tex]:
[tex]$$
M \approx \frac{31.5}{0.3816427360289367} \approx 82.5379262494639.
$$[/tex]
Thus, the monthly payment is approximately [tex]$82.54$[/tex].
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