Answer :
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
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
