# Thread: can you help me with this exercise?

1. ## can you help me with this exercise?

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. 2. Originally Posted by gusty 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$3. Hello, gusty! 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?
There is a formula for this very problem, but we can derive the formula ourselves
. . 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 and $20,000 is withdrawn.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 and$20,000 is withdrawn.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 and $20,000 is withdrawn.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 and$20,000 is withdrawn.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When the final $20,000 is withdrawn, the balance is zero.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$