Respuesta :
Answer:
Sean's balance at the end of ten years is $1,245.98.
Explanation:
The formula to compute the amount at the end of n years at r% interest rate is,
[tex]A=P[1+\frac{r}{100}]^{n}[/tex]
Here P = principal amount.
The amount deposited by Sean is, P₀ = $826.
For first n₀ = 3 years the annual effective interest rate is, r₀ = 3%.
⇒
Compute the amount in the bank account after 3 years as follows:
[tex]A_{0}=P_{0}[1+\frac{r_{0}}{100}]^{n_{0}}\\=826\times[1+\frac{3}{100} ]^{3}\\=826\times1.092727\\=902.593[/tex]
Amount at the end of 3 years is, $902.593.
Now this amount of P₁ = $902.593 is kept in the account at r₁ = 4% for n₁ = 2 years.
⇒
Compute the new amount in the bank account after total 5 years as follows:
[tex]A_{1}=A_{0}[1+\frac{r_{1}}{100}]^{n_{1}}\\=902.593\times[1+\frac{4}{100} ]^{2}\\=902.593\times1.0816\\=976.245[/tex]
Amount at the end of 5 years is, $976.245.
Now this amount of P₂ = $976.245 is kept in the account at r₂ = 5% for n₂₂ = 5 years.
⇒
Compute the new amount in the bank account after total 10 years as follows:
[tex]A_{2}=A_{1}[1+\frac{r_{2}}{100}]^{n_{2}}\\=976.245\times[1+\frac{5}{100} ]^{5}\\=976.245\times1.2763\\=1245.98[/tex]
Thus, Sean's balance at the end of ten years is $1,245.98.
