Can there be more than 5 positive integers such that they are in a harmonic progression ie there reciprocals are in arithmetic progression? If yes, the find some with more than 5 terms and show your working, and if not, prove your observation.

Note that \(\frac{1}{n!}, \frac{2}{n!}, \frac{3}{n!}, \ldots, \frac{n}{n!}\) form an arithmetic progression, and the numerator divides the denominator so they all simplify to unit fractions. Thus their reciprocals are positive integers that form a harmonic progression. Take \(n\) as large as you want.

Of course. Let \(a_1, a_2, \ldots, a_n\) be an arithmetic progression of positive integers, and let \(P\) be their least common multiple. Then \(\frac{kP}{a_1}, \frac{kP}{a_2}, \ldots, \frac{kP}{a_n}\) is a solution, for any positive integer \(k\). (The above is when you put \(a_i = i\) and \(k\) is such so \(kP = n!\).)

Whether the above gives all the solutions, I haven't proved it yet, although I think it does.

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TopNewestNote that \(\frac{1}{n!}, \frac{2}{n!}, \frac{3}{n!}, \ldots, \frac{n}{n!}\) form an arithmetic progression, and the numerator divides the denominator so they all simplify to unit fractions. Thus their reciprocals are positive integers that form a harmonic progression. Take \(n\) as large as you want.

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That's a cool solution. Do other solutions exist?

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Of course. Let \(a_1, a_2, \ldots, a_n\) be an arithmetic progression of positive integers, and let \(P\) be their least common multiple. Then \(\frac{kP}{a_1}, \frac{kP}{a_2}, \ldots, \frac{kP}{a_n}\) is a solution, for any positive integer \(k\). (The above is when you put \(a_i = i\) and \(k\) is such so \(kP = n!\).)

Whether the above gives all the solutions, I haven't proved it yet, although I think it does.

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