Terms of a Harmonic Progressions

If $a, b$ are two real numbers such that $a+ib \neq 0$ for all non-negative integer $i$ , then $\frac{1}{a}, \frac{1}{a+b}, \frac{1}{a+2b}, \ldots$ is a harmonic progression. Moreover, all harmonic progressions are in this form. Thus, if the $k$ -th term of a harmonic progression is $x$ , then there exists some $a,b$ such that $x = \frac{1}{a+(k-1)b}$ , or $(a+(k-1)b)x = 1$ .

The $m$ -th term is $n$ , so we have the equation $(a+(m-1)b)n = 1$ , or $bmn + (a-b)n = 1$ , or $(a-b)n = 1-bmn$ .

The $n$ -th term is $m$ , so we also have the equation $(a+(n-1)b)m = 1$ , or $bmn + (a-b)m = 1$ , or $(a-b)m = 1-bmn$ .

Thus $(a-b)n = (a-b)m$ , or $(a-b)(n-m) = 0$ . Since $n,m$ are distinct, $n-m \neq 0$ , so $a-b = 0$ or $a = b$ .

Substituting to either equation above, we obtain $(a+(m-1)a)n = 1$ , or $amn = 1$ . Thus $a = \frac{1}{mn}$ , and since $a = b$ , we have $b = \frac{1}{mn}$ .

Finally, the $mn$ -th term is given by $\frac{1}{a+(mn-1)b}$ . Substituting $a = b = \frac{1}{mn}$ gives $\frac{1}{\frac{1}{mn} + (mn-1) \cdot \frac{1}{mn}} = \frac{1}{mn \cdot \frac{1}{mn}} = \boxed{1}$

$\begin{cases} m^\text{th} \text{ term:} & \dfrac 1{a+(m-1)d} = n & \implies an + mnd - nd = 1 \\ n^\text{th} \text{ term:} & \dfrac 1{a+(n-1)d} = m & \implies am + mnd - md = 1 \end{cases}$

$\begin{aligned} \implies an + mnd - nd & = am + mnd - md \\ a(n-m) & = (n-m)d \\ \implies a & = d \end{aligned}$

$\begin{aligned} \implies k^\text{th} \text{ term:} \quad \frac 1{a+(k-1)d} & = \frac 1{a+(k-1)a} = \frac 1{ak} \end{aligned}$

$\begin{aligned} \implies m^\text{th} \text{ term:} \quad \frac 1{am} & = n \\ \implies amn & = 1 \end{aligned}$

$\begin{aligned} \implies (mn)^\text{th} \text{ term:} \quad \frac 1{amn} & = \frac 11 = \boxed{1} \end{aligned}$

To solve this problem, one only needs to plug in m=2 and n=1 and the answer follows immediately. However, the hard part is to prove that whatever m and n we choose, the (mn)th term is invariant.