1. **Problem statement:** Calculate the interest earned on 25000 invested for 7 years at 6% interest compounded in different ways. 2. **Formula for compound interest:** $$ A = P \left(1 + \frac{r}{n}\right)^{nt} $$ where $A$ is the amount, $P$ is the principal, $r$ is the annual interest rate (decimal), $n$ is the number of compounding periods per year, and $t$ is the time in years. 3. **Interest earned** is $I = A - P$. 4. **a. Annually ($n=1$):** $$ A = 25000 \left(1 + \frac{0.06}{1}\right)^{1 \times 7} = 25000 (1.06)^7 $$ Calculate: $$ (1.06)^7 = 1.503630 $$ $$ A = 25000 \times 1.503630 = 37590.75 $$ Interest: $$ I = 37590.75 - 25000 = 12590.75 $$ 5. **b. Semiannually ($n=2$):** $$ A = 25000 \left(1 + \frac{0.06}{2}\right)^{2 \times 7} = 25000 (1.03)^{14} $$ Calculate: $$ (1.03)^{14} = 1.5107 $$ $$ A = 25000 \times 1.5107 = 37767.50 $$ Interest: $$ I = 37767.50 - 25000 = 12767.50 $$ 6. **c. Quarterly ($n=4$):** $$ A = 25000 \left(1 + \frac{0.06}{4}\right)^{4 \times 7} = 25000 (1.015)^{28} $$ Calculate: $$ (1.015)^{28} = 1.5116 $$ $$ A = 25000 \times 1.5116 = 37790.00 $$ Interest: $$ I = 37790.00 - 25000 = 12790.00 $$ 7. **d. Monthly ($n=12$):** $$ A = 25000 \left(1 + \frac{0.06}{12}\right)^{12 \times 7} = 25000 (1.005)^{84} $$ Calculate: $$ (1.005)^{84} = 1.5132 $$ $$ A = 25000 \times 1.5132 = 37830.00 $$ Interest: $$ I = 37830.00 - 25000 = 12830.00 $$ 8. **e. Continuously compounded interest:** Formula: $$ A = P e^{rt} $$ Calculate: $$ A = 25000 e^{0.06 \times 7} = 25000 e^{0.42} $$ $$ e^{0.42} = 1.5210 $$ $$ A = 25000 \times 1.5210 = 38025.00 $$ Interest: $$ I = 38025.00 - 25000 = 13025.00 $$ **Final answers:** - Annually: 12590.75 - Semiannually: 12767.50 - Quarterly: 12790.00 - Monthly: 12830.00 - Continuously: 13025.00