Sequences

Package of problems

Solution

1. Can you find an arithmetic and a geometric sequence which each have second and fourth terms $u_2=6 \quad\text{and}\quad u_4=54 \text{ ?}$

Let’s start by looking for an arithmetic sequence. We are given the second and fourth terms so the common difference must be $d=\frac{1}{2}(54-6) = 24$ and the first term is $d$ less than the second, so $a=6-24 = -18.$

The arithmetic sequence is $(-18,6,30,54,78,...)$.

The geometric sequence has common ratio, $r$, such that $r^2=\frac{54}{6}=9 \quad\text{and so}\quad r=3.$ Its first term is $6\div3$ and the sequence is $(2,6,18,54,162,...)$.

We can illustrate these two sequences as sets of points on a graph.

We have drawn in lines on which the points lie. Can you describe the shape of the two lines?

Notice that when solving for $r$ above we could also have had $r=-3$. What would our geometric sequence have looked like then?

1. Can you find an arithmetic and a geometric sequence which each have $u_2=-2, \quad u_3=4 \quad\text{and}\quad u_5=16 \text{ ?}$

Having thought about how arithmetic and geometric sequences behave, your first reaction might be that it is impossible to have three terms in common. Let’s have a look at the numbers anyway…

Comparing $u_2$ and $u_3$, the arithmetic sequence must have $d=4-(-2)=6$ and this does fit with $u_5$. The first term, $a=u_2-d=-8$ and the arithmetic sequence is $(-8,-2,4,10,16,22,...)$.

For the geometric sequence, the common ratio must be $r=\frac{u_3}{u_2}=-2$. As a check, $\frac{u_5}{u_3}=4=r^2$ so this does work. The first term will be $\frac{u_2}{r}=1$ and the sequence is $(1,-2,4,-8,16,-32,...)$.

The reason this works is the negative common ratio. This is an example of an oscillating sequence where the values of the terms go above and below a central value, in this case zero.

As before, we can illustrate these sequences with a graph.

The terms of the arithmetic sequence lie along a straight line again, but this time we have had to draw a pair of exponential curves for the geometric sequence – one for the positive terms and one for the negative.

1. A geometric sequence $G$ has first term $8$ and common ratio $-\frac{1}{2}$. Can you find an arithmetic sequence that has the same first term and two other terms the same as the corresponding terms of $G$?

This is similar to the previous example in that $G$ has a negative common ratio and is therefore oscillating. Let’s write out the first few terms. $G=\left(8,-4,2,-1,\frac{1}{2},-\frac{1}{4},...\right)$

We want an arithmetic sequence that has $a=8$ and that meets $G$ again at two other terms. We could try matching some of the early terms and see what happens.

• Try matching the first two terms. Then we have $d=-12$ and the sequence would go $(8,-4,-16,-28,...)$. This is clearly not going to meet $G$ a third time.

• Try the first and third terms. $2d=-6$ so $d=-3$ and the sequence goes $(8,5,2,-1,-4,...)$. This is what we want because its first, third and fourth terms match with $G$.

What does this solution have in common with the solution to (2)? How are they related?

We have found a sequence with the first, third and fourth terms in common with $G$. Do you think that it is possible to find other arithmetic sequences with different sets of three terms in common with $G$?

1. Can you find an arithmetic and a geometric sequence that have four terms in common?

The only way we can get four terms of a geometric sequence to be linearly spaced is if all its terms are identical. A geometric sequence with common ratio $r=1$ and an arithmetic sequence with common difference $d=0$ will have identical terms if their first terms are the same.

You might argue that the sequence $(1,1,1,1,1,...)$ is neither arithmetic nor geometric, but it is generally considered to be both, as it satisfies the definition of both.