# a question on an arithmetic progression with four terms

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?

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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

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*}

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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 : $r^2=t$

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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.

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