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A loan for $8,000$ must be repaid with 6 year end payments at an annual rate of $11 \%$. What is the annual payment? I know that the present value of an annuity with end payments is $\frac{1-v^n}{i}$ where $v = 1/(1+i)$. Likewise, the future value is $(1+i)^{n}\frac{1-v^n}{i}$. How do I use this to solve this problem?

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If this is a homework problem, you may want to consider adding a homework tag. What is n in your problem? –  WWright Oct 15 '10 at 20:13
    
number of periods. –  PEV Oct 15 '10 at 20:18
    
Isn't it just that you want the present value of the annuity to equal the loan amount? Excel's PMT will let you check your answer –  Ross Millikan Oct 15 '10 at 20:23
    
How do you get the present value? I know $PV = \frac{FV}{(1+i)^n}$. But how yould you formulate this in terms of annuities? How do you do it by hand? –  PEV Oct 15 '10 at 20:30
    
@Trevor: Why the [statistics] tag? –  Américo Tavares Oct 16 '10 at 20:04
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up vote 2 down vote accepted

We have to find the value of constant (I assume, since nothing is specified on the contrary) payments $A$ given the principal $P$ during $n$ yearly periods and the interest rate $i$. The value of $A$ in the period $k$ is equivalent to the present value $A/\left( 1+i\right) ^{k}$ monetary units, where $i$ is the interest rate in each capitalization period. Summing in $k$, from 1 to $n$, we get the sum

$$\displaystyle\sum_{k=1}^{n}\dfrac{A}{\left( 1+i\right) ^{k}}$$

This is a geometric progression with ratio $r=1/(1+i)$ and first term $u_{1}=A/\left( 1+i\right)$ whose sum is:

$$\dfrac{A}{1+i}\dfrac{\left( \dfrac{1}{1+i}\right)^{n}-1}{\dfrac{1}{1+i}-1}=A\dfrac{\left( 1+i\right) ^{n}-1}{i\left( 1+i\right) ^{n}}=P.$$

Hence

$$A=P\dfrac{i\left( 1+i\right) ^{n}}{\left( 1+i\right) ^{n}-1}.$$

For the given problem, the payments will be made during $n=6$ years, with $i=11\%=\dfrac{11}{100}$ and $P=8000$:

$$A=8000\times \dfrac{0.11(1.11)^{6}}{(1.11)^{6}-1}\approx 1891$$


Sum of a geometric progression:

$$S=u_{1}+u_{2}+u_{3}+\ldots +u_{n}$$

$$rS=ru_{1}+ru_{2}+ru_{3}+\ldots +ru_{n-1}+ru_{n}$$

$$u_{k}=ru_{k-1}=u_{1}r^{k-1}$$

$$S-rS=\left( u_{1}+u_{2}+u_{3}+\ldots +u_{n}\right) -\left( ru_{1}+ru_{2}+ru_{3}+\ldots +ru_{n-1}+ru_{n}\right) $$

$$(1-r)S=u_{1}-ru_{n}$$

$$S=\dfrac{u_{1}-ru_{n}}{1-r}=\dfrac{u_{1}-u_{1}r^{n}}{1-r}=u_1\times\dfrac{1-r^{n}}{1-r}=u_1\times\dfrac{r^{n}-1}{r-1}.$$

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How did you sum the geometric series? I thought $\sum_{k=m}^{n} ar^{k} = \frac{a(r^{m}-r^{n+1})}{1-r}$. –  PEV Oct 16 '10 at 18:16
    
The sum of the geometric progression (starting at order $n=1$ until $n$) $u_{1}$, $u_{2}$, $\ldots$ ,$u_{n}$ is $S=u_{1}\times \dfrac{1-r^{n}}{1-r}$, where $r$ is the progression ratio. –  Américo Tavares Oct 16 '10 at 18:32
    
@Trevor: I added the derivation of the progression sum formula. –  Américo Tavares Oct 16 '10 at 18:49
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