Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

I am self-studying Discrete Mathemamatics and I've found the following example (in Portuguese).

Find an arithmetic progression with four terms $(a, a+r,a+2r,a+3r)$ such that the sum of its terms is $16$ and the product is $105$.

The author start by saying that we can assume that its terms are of the form $(y-3r,y-r,y+r,y+3r)$. I didn't understand that statement.

Here's what I did to prove his claim: The sum of its terms is equal to $4a+6r$. Then I put $y=2a+3r$ and then I get the arithmetic progression $(\dfrac{y-3r}{2},\dfrac{y-r}{2},\dfrac{y+r}{2},\dfrac{y+3r}{2})$. As you can see I was not able to prove his claim. What am I doing wrong?

I would appreciate your help.

share|improve this question
    
You have in fact proved his claim, but $y$ and $r$ need to be replaced by $y/2$ and $r/2$. It would have been less confusing if the author had chosen a different letter than $r$ in the second formula. –  Harald Hanche-Olsen Mar 14 '12 at 16:35
add comment

3 Answers 3

up vote 2 down vote accepted

You weren’t doing anything wrong: you just didn’t realize that the $r$ in the $y$ version is not the same quantity as the $r$ in the original $a$ version. Notice that the difference between consecutive terms in the $y$ version is $2r$, not $r$; thus, the new $r$ must be just half as big as the original one. How did he know that the original $r$ was even?

The product is odd, so all four terms must be odd. This means that the original $r$ must be even. Let $r=2s$ for some integer $s$. Then the original four terms are $a,a+2s,a+4s$, and $a+6s$. Now let $y=a+3s$; then you can rewrite these four terms as $y-3s,y-s,y+s$, and $y+3s$. Now just change the name from $s$ to $r$, and you have the terms written as $y-3r,y-r,y+r$, and $y+3r$, though this $r$ is half of the original one.

The reason for rewriting the progression as the author did is that it makes it gives the sum and product relatively nice forms: the sum is just $4y$, and the product is $$\begin{align*}(y-r)(y+r)(y-3r)(y+3r)&=(y^2-r^2)(y^2-9r^2)\\ &=y^4-10r^2y^2+9r^4\;. \end{align*}$$

share|improve this answer
add comment

A bit of motivation for this notation:

  • usually, an arithmetic progression using three terms is written as $a-r,a,a+r$. Why is that? Because if you add or multiply the terms of the arithmetic progression you get simpler expression than almost every other notation, like $b,b+q,b+2q$, etc.

  • the same reasoning is for the four term progression. Writing it as $b,b+q,b+2q,b+3q$ doesn't help you very much when you calculate the product. You would need some kind of symmetry to be able to use the formula $(a-b)(a+b)=a^2-b^2$, which simplifies calculations. This symmetry can be only with respect to $r$, which is the arithmetic mean of the four terms.

share|improve this answer
add comment

So ,you know that :

$4a+6r=16 \Rightarrow a=\frac{8-3r}{2}$

$a(a+r)(a+2r)(a+3r)=105$

If you substitute $a$ into second equation you should get following equality :

$$\frac{9}{16}r^4-40r^2+151=0$$

This equation you can solve using substitution : $r^2=t$

share|improve this answer
add comment

Your Answer

 
discard

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.