Question

Karen wants to invest $750 per quarter for a period of 25 years in order to accumulate a value of $120,000. What minimum annual rate (compunded quarterly) would Karen need to receive to reach her goal?

Answer 1

Answer:

Step-by-step explanation:

For Compound Interest with continuous deposit,

We have that:

$\frac{dS}{dt}=rS+k$

$\frac{dS}{dt}-rS=k$

Using Integrating factor: $e^{\int\ {-r} \, dt }$=$e^{-rt}$

$\frac{dS}{dt}e^{-rt}=ke^{-rt}$

Taking Integrals from 0 to t

$Se^{-rt}=\frac{k}{-r}e^{-rt}+C\\S=\frac{k}{-r}+Ce^{rt}\\When t=0\\C=S_{0}+\frac{k}{r}$

Now substituting back C

We have:

$Se^{-rt}=\frac{k}{-r}e^{-rt}+S_{0}+\frac{k}{r}\\S(t)=S_{0}e^{rt}+\frac{k}{r}-\frac{k}{r}e^{rt}\\S(t)=S_{0}e^{rt}+\frac{k}{r}(1-e^{rt})$

$Se^{-rt}=\frac{k}{-r}e^{-rt}+S_{0}+\frac{k}{r}\\S(t)=S_{0}e^{rt}+\frac{k}{r}-\frac{k}{r}e^{rt}\\S(t)=S_{0}e^{rt}+\frac{k}{r}(1-e^{rt})$

If initial deposit =$750, Continuous Deposit = 750, t= 4X 25 years=100

$S(t)=S_{0}e^{rt}+\frac{k}{r}(1-e^{rt})\\120000=750e^{100r}+\frac{750}{r}(1-e^{100r})\\160=e^{100r}+\frac{750}{r}-\frac{750}{r}e^{100r}$