# Exploring Sequences and Lines

Here’s another favorite problem that could be used for any middle or high school students who’ve been exposed to both arithmetic sequences and linear equations.

There is a family of lines, $Ax+By+C=0$, for which $\left\{ A, B, C \right\}$ can be any arithmetic sequence.  What do all members of this family have in common?

As with most great math problems, the problem is simply stated and can be approached from many different perspectives.  In the end, this one isn’t that difficult to crack, but the solution (at least to me) is not intuitively obvious from the problem statement. If you’ve not seen this before, please try it before reading further.

WARNING:  Problem Solution Follows

MOST COMMON STUDENT APPROACH:  Given the problem’s phrasing, most begin by writing out a few possible equations using different arithmetic sequences, typically with positive common differences.  After a few tries, most get a little frustrated as no obvious pattern emerges from the equations.

It is this point that is typically the most difficult for me as a teacher.  I want to help anyone who struggles, but “help” often means allowing others the room to struggle and to find ways of solving problems on their own.  Only when a student or group gets really frustrated do I sometimes ask, “Are there any other ways you can look at this problem or at your work?”

Eventually, most decide to graph their equations to see if anything pops out.  The following graph was submitted this past semester by one of my precalculus students using the free online Desmos calculator.

Two lines intersecting in a point is common.  Three or more in a single point almost always indicates something interesting.  Seven lines through a single point is screaming for attention!  From this graph, all lines in this family apparently contain the point (1,-2).  That seems a bit random until you investigate further, but pattern recognition is just half of the battle for a mathematician.  Now that something interesting has been discovered, a mathematician wants to know if this is a fluke or something inherent to all members of the family.

NOTE:  All graphs my students have produced over the years have always contained specific equations.  I don’t know that it’s any more enlightening, but I did create on Desmos a slider version of the graphs of this family with one slider for the initial term (A) and a second for its common difference (d).

UNIQUE SOLUTION METHODS FROM STUDENTS:

All successful solutions eventually rewrite the sequence $\left\{ A,B,C \right\}$ to $\left\{ A,A+d,A+2d \right\}$ where d is the common difference for a generic arithmetic sequence with initial term A.

Method I: After discovering the common point of intersection, most plug it into the left side of the equation and simplify to get

$Ax+By+C=A\cdot 1+\left( A+d\right)\cdot -2+\left( A+2d \right)=0$.

Because the left side reduces to zero for all generic arithmetic sequences, $\left\{ A,A+d,A+2d \right\}$, (1,-2) must be common to all members of this family.

A few students aren’t comfortable dealing with 0=0, so these tend to plug in $x=1$ and solve for y to get $y=-2$, proving that the y-coordinate for $x=1$ for all members of this family is always -2.

Method II:  A few students every year work algebraically from properties of arithmetic sequences.  For any arithmetic sequence, $\left\{ A,B,C \right\}$, $\frac{A+C}{2}=B$.  This rewrites to $1\cdot A-2\cdot B+C=0$, so whenever $\left( x,y \right)=\left(1,-2 \right)$, then $Ax+By+C=0$ is a fundamental property of all arithmetic sequences.

Personally, I think this method gets closest to explaining why the point (1,-2) is the common characteristic of this family.

Method III:  This year, I had a student take an approach I’d never seen before.  She defined one arithmetic sequence as $\left\{ a,a+d,a+2d \right\}$ and another as $\left\{ m,m+n,m+2n \right\}$ for any real values of a,d,m, and n.  This leads to a system of equations: $a\cdot x+(a+d)\cdot y+(a+2d)=0$ and $m\cdot x+(m+n)\cdot y+(m+2n)=0$ .  If you have some younger students or if all the variables make you nervous, the solution is available via Wolfram Alpha.

Still, this system is pretty easy to manipulate.  If you multiply the first equation by m and the second by a, the x-terms will eliminate with subtraction, giving

$m\cdot((a+d)\cdot y+(a+2d))-a\cdot((m+n)\cdot y+(m+2n))=0$.

Solving for y shows that all of the coefficients simplify surprisingly easily.

$((ma+md)-(am+an))\cdot y=-(ma+2md)+(am+2an)$
$(md-an)\cdot y = -2\cdot (md-an)\Longrightarrow y=-2$

From here, determining $x=1$ is easy, proving the relationship.

SOLUTIONS and APPROACHES NOT YET OFFERED BY STUDENTS:

Approach A:  High school students don’t often think about ways to simplify problem situations, especially at the beginning of problems.  One way I did that for this problem in later class discussions was to recognize that it one of the terms in the arithmetic sequence was 0, you didn’t need to deal with nearly as many terms. For example, if your sequence was ${1,0,-1}$, the linear equation would be $x-1=0$.  Similarly, the sequence $\left\{ 01,2 \right\}$ leads to $y+2=0$.  Obviously, the only thing these two lines have in common is the point (1,-2).  A proof of the property must still be established, but this is one of the fastest ways I’ve seen to identify the central property.

Approach B:  A purely algebraic approach to this problem could redefine the arithmetic sequence as $\left\{ a,a+d,a+2d\right\}$as before, giving:

$a\cdot x+(a+d)\cdot y+(a+2d)=0$

Collecting like terms gives

$(x+y+1)\cdot a+(y+2)\cdot d=0$.

The values of a and d must remain as parameters to include all possible arithmetic sequences.  Because the equation always equals 0, the coefficients of a and d are both 0, making $y=-2$ (for the coefficient of d) and therefore $x=1$.

EXTENSION:

We once had a test question at the end of the unit containing this exercise.  Basically, it reminded students that they had discovered that all lines $Ax+By+C=0$, for which $\left\{ A, B, C \right\}$ can be any arithmetic sequence contained the point (1,-2).  It then asked for an equation of a family of linear functions using the same arithmetic $\left\{ A, B, C \right\}$ that all contained the point (1,2).

The two most common responses we’ve seen involve a reflection or a vertical translation.  (1,-2) can become (1,2) by reflecting over the x-axis, so making the y-values negative would do the trick:  $Ax-By+C=0$.  Similarly, (1,-2) can become (1,2) by translating up 4 units, giving $Ax+B(y-4)+C=0$.