# Suppose $10,000 is invested at an annual rate of displaystyle{2.4}% for 10 years. Find the future value if interest is compounded as follows Joni Kenny 2021-02-12 Answered Suppose$10,000 is invested at an annual rate of $$\displaystyle{2.4}\%$$ for 10 years. Find the future value if interest is compounded as follows

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Amount for compound interest:
If P amount of money is invested for t years at r% rate of interest which compounded n times annually then total amount of money to be received is calculated as per the formula given below:
$$\displaystyle{A}={P}{\left({1}+\frac{r}{{n}}\right)}^{r}{t}$$
(a)
It is given that the principal amount of money is $10,000, rate of interest is $$\displaystyle{2.4}\%$$, time is 10 years and the sum is compounded annually. This implies that the value of P is 10000, value of r is 0.024, value of t is 10 and value of n is 1. $$\displaystyle{A}={10000}{\left({1}+\frac{0.024}{{1}}\right)}^{{{1}{\left({10}\right)}}}$$ $$\displaystyle={10000}{\left({1}+{0.024}\right)}^{10}$$ $$\displaystyle\approx{12676.51}$$ Thus, is the future value will be$12,676.51 in the account after 10 years
(b)
It is given that the principal amount of money is $10,000, rate of interest is $$\displaystyle{2.4}\%$$, time is 10 years and the sum is compounded annually. This implies that the value of P is 10000, value of r is 0.024, value of t is 10 and value of n is 4. $$\displaystyle{A}={10000}{\left({1}+\frac{0.024}{{4}}\right)}^{{{4}{\left({10}\right)}}}$$ $$\displaystyle={10000}{\left({1}+\frac{0.024}{{4}}\right)}{40}$$ $$\displaystyle\approx{12703.38}$$ Thus, is the future value will be$12703.37 in the account after 10 years
(c)
It is given that the principal amount of money is $10,000, rate of interest is $$\displaystyle{2.4}\%$$, time is 10 years and the sum is compounded annually. This implies that the value of P is 10000, value of r is 0.024, value of t is 10 and value of n is 12. $$\displaystyle{A}={10000}{\left({1}+\frac{{{0.024}}}{{12}}\right)}^{{{12}{\left({10}\right)}}}$$ $$\displaystyle={10000}{\left({1}+\frac{{{0.024}}}{{12}}\right)}{120}$$ $$\displaystyle\approx{12709.44}$$ Thus is the future value will be$12,709.44 in the account after 10 years
(d) It is given that the principal amount of money is $10,000, rate of interest is $$\displaystyle{2.4}\%$$, time is 10 years and the sum is compounded annually. This implies that the value of P is 10000, value of r is 0.024, value of t is 10 and value of n is 365. $$\displaystyle{A}={10000}{\left({1}+\frac{{{0.024}}}{{365}}\right)}^{{{365}{\left({10}\right)}}}$$ $$\displaystyle={10000}{\left({1}+\frac{{{0.024}}}{{365}}\right)}{3650}$$ $$\displaystyle\approx{12712.39}$$ Thus is the future value will be$12712.39 in the account after 10 years
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