Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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?

share|cite|improve this question
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
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:

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


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

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:

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

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


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



share|cite|improve this answer
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

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.