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# A note on geometric sequences

We’re s--o excited about this!!! As in, EXCITED!!! Indeed, this is expected, considering what is to be expounded: “any” geometric sequence can be encapsulated as slopes of a simple triangle like the one below:

And the procedure is easy: first find a triangle with slopes , then the remaining terms can be obtained as slopes of line segments drawn from a fixed vertex to a fixed side .

## Positive common ratio

In our first three examples, we’ll generate the entire geometric sequence

that is, (positive) powers of .

#### Example

Find coordinates for the vertices of with slopes .

There are different ways to obtain a triangle whose slopes are . One can begin with an arbitrary point . Then the two other points can be chosen as and . With these, the slopes of sides will be , respectively.

In the present case, let’s first choose as , for simplicity. Since the sequence is , we have and . So and .

The entire geometric sequence can be obtained from the above triangle.

#### Example

For each , PROVE that lies within , where is the point and is the point .

Note that is a point on . Let’s use distances to show that it’s indeed internal. We prove that .

This proves that for each , the point lies internally on .

#### Example

For each , let . PROVE that the slope of is , where is the point .

This follows by direct calculation:

Thus, when , the corresponding slopes will be ; that is, .

By defining and using with vertices at , we’ve seen that the entire geometric progression can be obtained as slopes of the line segments , together with with slope . Notice that when , becomes and when , becomes . What happens to when ?

Put . Then has coordinates . Together with , the slope of line segment is

as expected.

Put , then has coordinates . Together with , the slope of line segment is

as expected.

Put , then has coordinates . Together with , the slope of line segment is

as expected.

Three things to note from the above calculation:

• the points are defined relative to the chosen coordinates for ;
• the line segments that describe the geometric progression do not get to the midpoint of side ;
• the roles of vertex and side cannot be altered; doing so will distort the orderly arrangement of the line segments. See the next two examples.

#### Example

Given with vertices , find coordinates for a point on such that the slope of the line segment is .

Let’s find the equations of and and then solve the resulting linear system:

Thus, is the point . It is external to side — we don’t want that.

#### Example

Given with vertices , find coordinates for a point on such that the slope of the line segment is .

Let’s find the equations of and and then solve the resulting linear system:

Thus, is the point . It is external to side — we don’t want that.

In the next two examples, we generate the geometric progression

#### Example

Find coordinates for the vertices of whose side slopes are .

Take . Then the slope of is , the slope of is , and the slope of is .

#### Example

Let . For each , PROVE that the slope of the line segment is , where is the point .

Observe that each lies within (). By direct calculation:

Thus, we obtain the geometric progression by putting . Together with the first three terms that are the slopes of the original triangle, we obtain the entire geometric progression .

Now to the positive powers of ; that is, the geometric progression

#### Example

Explain how to obtain the entire geometric progression as slopes.

Easy-peasy.

Begin with a in which the slopes of the sides are . To this end, take . Then the slope of is , the slope of is , and the slope of is .

Next, let . Then, by direct calculation (as in and ), the slope of is , for . Therefore, letting successively, we obtain the terms . If we now append the slopes of the sides of (which are ), we obtain the entire geometric progression .

## Reciprocal terms

In addition to obtaining , we also get , once we go past a certain point; let’s call it a terminal point (see the exercises at the end ).

#### Example

Given with vertices , find coordinates for a point on such that the slope of is .

In Example we had . Put , then . Take . Then is an internal point on ; further, the slope of is

as desired. Similarly, by putting , we obtain the terms . Thus, inside a simple, single triangle lies the infinite geometric progression

disguised as slopes.

#### Example

Given with vertices , let . PROVE that .

Put , then . Put , then , as also found in a previous example. By the distance formula:

Observe that the slopes of the sides of are ; they form a geometric progression with a common ratio of , which is the square of the common ratio of the geometric progression formed by the slopes of the sides of the parent .

## Takeaway

Slopes are usually introduced sometime in grade (or earlier), while geometric sequences are usually studied in grade (or earlier). Here, we’ve managed to show a close connection between the two concepts, using the medium of a triangle.

Throughout we worked with positive integer common ratios. But the common ratios can also be negative, and in such cases things become much more exciting.

1. (Depressed cubic) Consider with vertices . PROVE that:
• the foot of the altitude from vertex is ;
• the length of the altitude from vertex is ;
• the area of is (this becomes a depressed cubic if written in the form . Moreover, if is an integer ( of course), then this area is always a positive integer).
2. Let . PROVE that as .
3. Let . PROVE that as .
4. (Special points) Given with vertices , PROVE that there are (internal) points and on such that:
• the slope of is undefined and the slope of is zero ( for vertical, for horizontal);
• and trisect side ();
• the areas of are sq. unit each (making the overall area of to be sq. units).
5. (Higher powers) Given with vertices at , the slopes of its sides form a geometric progression with common ratio . Find two points and on such that the slopes of the sides of form a geometric progression with common ratio .
6. (Negative region) Given with vertices , find two points and on such that the slope of any line segment , where is a point between and , is negative.
7. (Base five) Explain how to generate the geometric progression using slopes.
8. (Base six) Explain how to generate the geometric progression using slopes.
9. (Base seven) Explain how to generate the geometric progression using slopes.
10. (Terminal point) Given with vertices , determine the coordinates of the terminal point on (a point on such that the slope of the line segment is ; this is also where the slopes of the line segments from vertex change from positive powers of to negative powers of ).