How much to deposit today

brunettie91

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Joined
Sep 4, 2013
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I have no idea how to do this one...

The amount a person would need to deposit today to be able to withdraw $6000 each year for 10 years from an account earning 6% interest.

I have a chart with all the compounding interest..I just don't know how to start the equation or what equation to use
 
Hello, brunettie91!

This is more involved than Compound Interest.


The amount a person would need to deposit today to be able to withdraw $6000 each year
for 10 years from an account earning 6% interest.

Amortization Formula: .A  =  Pi(1+i)n(1+i)n1\displaystyle A \;=\;P\dfrac{i(1+i)^n}{(1+i)^n-1}

. . where: .{A=periodic paymentP=principal borrowedi=periodic interest raten=number of periods}\displaystyle \begin{Bmatrix}A &=&\text{periodic payment} \\ P &=& \text{principal borrowed} \\ i &=& \text{periodic interest rate} \\ n &=& \text{number of periods} \end{Bmatrix}


Consider that you borrowed P\displaystyle P dollars.
The loan charges 6% interest annualy.
You will repay the loan with annual payments of $6000 each for ten years.
How much can you borrow?

We have: .(A=6000i=0.06n=10)\displaystyle \begin{pmatrix}A &=& 6000 \\ i &=& 0.06 \\ n &=& 10\end{pmatrix}

Substitute: .\(\displaystyle 6000 \:=\:p\dfrac{(0.06)(1.06)^{10}}{(1.06)^{10} - 1} \quad\Rightarrow\quad 6000 \:=\:p(0.135867958)\)

Hence: .P  =  60000.135867958  =  44, ⁣160.52231\displaystyle P \;=\;\dfrac{6000}{0.135867958} \;=\;44,\!160.52231


Therefore: .P    $44,160.52\displaystyle P \;\approx\;\$44,160.52
 
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