Things you might have noticed

Write down three numbers. Can you find an arithmetic progression containing all three of these numbers (in ANY order)?

Is it always possible to find an arithmetic progression containing any three numbers chosen (in ANY order)?

We will consider these questions in the context of the cards given in the Suggestion.

Cards C, D, E and I

\(4\), \(8\), \(14\)

\(78\), \(39\), \(169\)

\(-11\), \(-3\), \(19\)

\(2\), \(5\), \(5\)

If we want them to belong to an arithmetic progression (from now on referred to as an AP), we need to identify a common difference for each set of three numbers.

  • How might you do this for card C?
  • Can the same approach be applied to cards D, E & I?
  • Is it possible to find an AP that contains the numbers in card D in their given order?
  • What is different about card I? Is there ever an occasion where an AP could contain a repeated value?

Let’s say we suggest the following AP for card C, \[2, 4, 6, 8, \dots\] Are there any other, different APs that could have been suggested?

Cards F and G

\(\frac{1}{2}\), \(\pi\), \(5\pi - 2\)

\(1\), \(\frac{\pi}{2}\), \(\pi\)

We might choose to look at card F first. Beginning as for cards C, D and E, by looking for a common difference we find that the difference between the first two values is \(\pi-\frac{1}{2}\) and the second two values is \(4\pi - 2\). Now, for us to find an AP we require that \[p\left(\pi - \frac{1}{2}\right) = q(4\pi - 2)\] where \(p\) and \(q\) are integers.

Can you see any values for \(p\) and \(q\) that could work here?

  • What about if \(q=1\) (i.e. \(4\pi - 2\) is a multiple of \(\pi - \frac{1}{2}\))?

Now let’s think about card G. As before, looking for a common difference we find that the difference between the first two values is \(\frac{\pi}{2} - 1\) and the second two values is \(\frac{\pi}{2}\). But what do we do with this information now?

As above, if we have an AP then we can write \[p\left(\frac{\pi}{2} - 1\right) = q\left(\frac{\pi}{2}\right)\] where \(p\) and \(q\) are integers.

Expanding the brackets and rearranging we have \[\pi=\frac{2p}{p-q}.\]

Why did we choose to rearrange the equation to make \(\pi\) the subject?

  • Why might this be useful? What do we know about \(\pi\)?

Could \(p\) ever be equal to \(q\)?

  • What would this mean for the progression?
  • What would it mean algebraically?

\[\pi=\frac{2p}{p-q}\]

As \(\pi\) is irrational and all the other terms in the expression are integers, this is impossible and we have reached a contradiction. So it is NOT possible to find an AP containing all three of the terms on card G.

Cards A, B and H

\(\sqrt{2}\), \(\sqrt{3}\), \(\sqrt{5}\)

\(1\), \(\sqrt{2}\), \(2\)

\(4\sqrt{3} - 5\), \(-3\), \(-6\sqrt{3}\)

Based on your thinking so far, or otherwise, do you have any feeling about whether or not it will be possible to find APs for the numbers on these cards?

Below is a way to think about card A. It proves that it is impossible to find an AP containing all three of the values.

Does this mean that it is impossible to find an AP for cards B and H too?

The difference between the first two values is \(\sqrt{3} - \sqrt{2}\) and between the second two values is \(\sqrt{5} - \sqrt{3}\).

As for cards F and G, if we have an AP then we can write \[p\left(\sqrt{3} - \sqrt{2}\right) = q\left(\sqrt{5} - \sqrt{3}\right)\] where \(p\) and \(q\) are integers.

There are not ‘obvious’ values of \(p\) and \(q\) that will work here so let’s continue as we did for card G.

Expanding the expression and rearranging we have \[(p+q)\sqrt{3} = p\sqrt{2} + q\sqrt{5}.\]

At this point we have three different irrational terms in our equation. This is different to what happened with card G.

What happens if we square both sides?

Squaring both sides we get \[3(p+q)^2 = 2p^2 + 5q^2 + (p+q)\sqrt{10}.\]

Rearranging to make \(\sqrt{10}\) the subject gives us \[\sqrt{10} = \frac{3(p+q)^2 - 2p^2 + 5q^2}{p+q}.\] As \(\sqrt{10}\) is irrational, but all of the terms on the right-hand side are integers, we have a contradiction. So it is NOT possible to find an AP containing all three of the values on card A.

You might like to think about cards B and H in a similar way before looking at the suggestions below for card H.

The difference between the first two values is \(4\sqrt{3}-2\) and between the second two values is \(6\sqrt{3}-3\).

As previously, if we have an AP then we can write \[p\left(4\sqrt{3}-2\right) = q\left(6\sqrt{3}-3\right)\] where \(p\) and \(q\) are integers.

We might now notice that the expressions within the brackets here can be factorised to give \[2p\left(2\sqrt{3}-1\right) = 3q\left(2\sqrt{3}-1\right)\] and then simplified to show that \[2p=3q.\]

This tells us that we can find values of \(p\) and \(q\), and therefore that we can identify a common difference and then an AP that contains the values on this card.

What might have happened if we had continued without factorising the expression \[p\left(4\sqrt{3}-2\right) = q\left(6\sqrt{3}-3\right)?\]

If we had continued as for card A, we might have expanded the brackets and rearranged to get \[\left(4p - 6q\right)\sqrt{3} = 2p - 3q\] but this suggests that \[\sqrt{3} = \frac{2p-3q}{2(2p-3q)} = \frac{1}{2}.\] This is a contradiction and suggests that we cannot find an AP containing the values on this card.

Can you identify what has gone wrong with our line of reasoning here?

Card Is it possible to find an AP? A possible AP
A No
B No
C Yes \(2, 4, 6, 8, \dots\)
D Yes \(13, 26, 39, 52, \dots\)
E Yes \(-11,-9,-7,-5,\dots\)
F Yes \(\left(\frac{1}{2}\right), \pi, \left(2\pi-\frac{1}{2}\right), \left(3\pi - 1\right), \dots\)
G No
H Yes \(\left(4 \sqrt{3}-5\right), \left(2 \sqrt{3} - 4\right), -3, \left(-2 \sqrt{3} - 2\right), \dots\)
I No