how much I must deposit now in a 12% account (annual interest) if I want to be able to withdraw $20 000 at the end of each of the next five years.
are we doing simple interest here?
If so, use $\displaystyle I = PRT$
where $\displaystyle I$ is the simple interest, $\displaystyle P$ is the principal, $\displaystyle R$ is the rate of interest, and $\displaystyle T$ is the time
at the end of 5 years, $\displaystyle I = P(0.12)(5) = 0.6P$
so, at the end, we want to have the amount we invested, $\displaystyle P$, with the interest, $\displaystyle I$, to be $20,000
hence we want, $\displaystyle P + I = 20000$
Hello, gusty!
There is a formula for this very problem, but we can derive the formula ourselvesHow much I must deposit now in a 12% account (annual interest)
if I want to be able to withdraw $20,000 at the end of each of the next five years?
. . if we're willing to invest some time and effort.
Let $\displaystyle P$ = amount of our initial deposit.
End of year-1: its value grows to $\displaystyle 1.12P$ dollars,
. . and $20,000 is withdrawn.
The balance is: .$\displaystyle 1.12P - 20,\!000$
End of year-2: the value grows to $\displaystyle (1.12)(1.12P - 20,\!000)$,
. . and $20,000 is withdrawn.
The balance is: .$\displaystyle 1.12^2P - 1.12(20,\!000) -20,\!000$
End of year-3: the value grows to $\displaystyle 1.12\left[1.12^2P - 1.12(20,\!000) - 20,\!000\right]$
. . and $20,000 is withdrawn.
The balance is: .$\displaystyle 1.12^3P - 1.12^2(20,\!000) - 1.12(20,\!000) - 20,\!000 $
End of year-4: the value grows to $\displaystyle 1.12\left[1.12^3P - 1.12^2(20,\!000) - 1.12(20,\!000) - 20,\!000\right] $
. . and $20,000 is withdrawn.
The balance is: .$\displaystyle 1.12^4P - 1.12^3(20,\!000) - 1.12^2(20,\!000) - 1.12(20,\!000) - 20,\!000 $
End of year-5: the value grows to $\displaystyle 1.12\left[1.12^4 - 1.12^3(20,\!000) - 1.12^2(20,\!000) - 1.12(20,\!000) - 20,\!000\right]$
When the final $20,000 is withdrawn, the balance is zero.
. . $\displaystyle 1.12^5P - 20,\!000(1.12^4) - 20,\!000(1.12^3) - 20,\!000(1.12^2) - 20,\!000(1.12) - 20,\!000 \;=\;0$
. . $\displaystyle 1.12^5P \;=\;20,\!000\underbrace{\bigg[1.12^4 + 1.12^3 + 1.12^2 + 1.12 + 1\bigg]}_{\text{Geometric series}} $ .[1]
On the right, we have a geometric series with: .$\displaystyle a = 1,\;r = 1.12,\;n = 5$
. . Its sum is: .$\displaystyle S \:=\:\frac{1.12^5-1}{1.12-1} $
Substitute into [1]: .$\displaystyle 1.12^5P \;=\;20,\!000\,\frac{1.12^5-1}{0.12} \quad\Rightarrow\quad P \;=\;20,\!000\,\frac{1.12^5-1}{0.12(1.12^5)}$
Therefore: .$\displaystyle P \;=\;72,\!095.52405 \;\approx\;\$72,\!095.52$