4
$\begingroup$

A sequence: 1/2, 1/3, 1/4, 1/5, ......
Within it, Does there exist an arithmetic progression of five fractions.
And is there an arithmetic progression with more than five fractions from this sequence.

I found 1/2, 1/3, 1/6 is an arithmetic progression,
then I halved them to get 1/4, 1/6, 1/12, and add 1/3 to the front to make an arithmetic progression of four fractions: 1/3, 1/4, 1/6, 1/12.
From here, I divided the four fractions by 5 to get: 1/15, 1/20, 1/30, 1/60, and add 1/12 to the front. Then I have an arithmetic progression of five fractions: 1/12, 1/15, 1/20, 1/30, 1/60.

But I do not know if there is an arithmetic progression with more than five fractions and how to make them.

$\endgroup$
9
$\begingroup$

Hint: The arithmetic progressions you found written in ascending order are:

$\dfrac{1}{6}, \dfrac{2}{6}, \dfrac{3}{6}$

$\dfrac{1}{12}, \dfrac{2}{12}, \dfrac{3}{12}, \dfrac{4}{12}$

$\dfrac{1}{60}, \dfrac{2}{60}, \dfrac{3}{60}, \dfrac{4}{60}, \dfrac{5}{60}$

See if you can generalize this. Spoiler below.

For any integer $n$, let $L_n = \text{lcm}(1,2,\ldots,n)$. Then, the numbers $\dfrac{1}{L_n}, \dfrac{2}{L_n}, \ldots, \dfrac{n}{L_n}$ are an arithmetic progression and are all in the sequence since $\dfrac{L_n}{n}$ is a positive integer. You could also use $\dfrac{1}{n!}, \dfrac{2}{n!}, \ldots, \dfrac{n}{n!}$.

$\endgroup$
2
$\begingroup$

Let $\frac1{a_1}, \frac1{a_2},\ldots ,\frac1{a_n}$ be a decreasing arithmetic progression of unit fractions of length $n$. Then you can add a rational number at the front, to make the arithmetic progression $$\frac k{a_0}, \frac1{a_1}, \frac1{a_2}, \ldots , \frac1{a_n}$$with $k, a_0$ being coprime integers. Now divide all of them by $k$, and you have a new arithmetic progression of unit fractions of length $n+1$. This you can keep going indefinitely.

You are, of course, free to divide by any multiple of $k$, but apart from that everything is forced, and given two fractions to start, whatever result you get will be similar (in the geometric sense) to what you get by dividing with exactly $k$. So essentially, the sequences you end up with are entirely decided by what two fractions you start with.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.