Absolutely, let's go through the solution step-by-step.
Given:
- Principal (P) = \[tex]$10,000
- Annual interest rate (r) = 6.5% = 0.065
- Time (t) = 3 years
We need to find the accumulated value using different compounding methods:
### a. Compounded Semiannually
For semiannual compounding, the number of compounding periods per year (n) is 2.
Using the formula \( A = P\left(1+\frac{r}{n}\right)^{nt} \), we have:
- \( P = 10000 \)
- \( r = 0.065 \)
- \( n = 2 \)
- \( t = 3 \)
Substituting these values, we get:
\[
A_{\text{semiannual}} = 10000 \left(1+\frac{0.065}{2}\right)^{2 \times 3}
\]
\[
A_{\text{semiannual}} = 10000 \left(1+0.0325\right)^6
\]
\[
A_{\text{semiannual}} = 10000 (1.0325)^6
\]
Calculating the result:
\[
A_{\text{semiannual}} \approx 10000 \times 1.211546
\]
Rounding to the nearest cent:
\[
A_{\text{semiannual}} = 12115.47
\]
### b. Compounded Quarterly
For quarterly compounding, the number of compounding periods per year (n) is 4.
Using the formula \( A = P\left(1+\frac{r}{n}\right)^{nt} \), we have:
- \( P = 10000 \)
- \( r = 0.065 \)
- \( n = 4 \)
- \( t = 3 \)
Substituting these values, we get:
\[
A_{\text{quarterly}} = 10000 \left(1+\frac{0.065}{4}\right)^{4 \times 3}
\]
\[
A_{\text{quarterly}} = 10000 \left(1+0.01625\right)^{12}
\]
\[
A_{\text{quarterly}} = 10000 (1.01625)^{12}
\]
Calculating the result:
\[
A_{\text{quarterly}} \approx 10000 \times 1.213408
\]
Rounding to the nearest cent:
\[
A_{\text{quarterly}} = 12134.08
\]
### c. Compounded Monthly
For monthly compounding, the number of compounding periods per year (n) is 12.
Using the formula \( A = P\left(1+\frac{r}{n}\right)^{nt} \), we have:
- \( P = 10000 \)
- \( r = 0.065 \)
- \( n = 12 \)
- \( t = 3 \)
Substituting these values, we get:
\[
A_{\text{monthly}} = 10000 \left(1+\frac{0.065}{12}\right)^{12 \times 3}
\]
\[
A_{\text{monthly}} = 10000 \left(1+0.00541667\right)^{36}
\]
\[
A_{\text{monthly}} = 10000 (1.00541667)^{36}
\]
Calculating the result:
\[
A_{\text{monthly}} \approx 10000 \times 1.214672
\]
Rounding to the nearest cent:
\[
A_{\text{monthly}} = 12146.72
\]
### Compounded Continuously
For continuous compounding, we use the formula \( A = P e^{rt} \).
Using this, we have:
- \( P = 10000 \)
- \( r = 0.065 \)
- \( t = 3 \)
Substituting these values, we get:
\[
A_{\text{continuously}} = 10000 \times e^{0.065 \times 3}
\]
\[
A_{\text{continuously}} = 10000 \times e^{0.195}
\]
Calculating the result:
\[
A_{\text{continuously}} \approx 10000 \times 1.215311
\]
Rounding to the nearest cent:
\[
A_{\text{continuously}} = 12153.11
\]
### Summary
Thus, the accumulated values for the various compounding methods are:
a. Compounded Semiannually: \$[/tex]12,115.47
b. Compounded Quarterly: \[tex]$12,134.08
c. Compounded Monthly: \$[/tex]12,146.72
d. Compounded Continuously: \$12,153.11
