- 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?
- 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.
- 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 \(G\)?
- 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.
