
5. Three nonzero real numbers $a, b, c$ are said to be in harmonic progression if $\frac{1}{a}+\frac{1}{c}=\frac{2}{b}$. Find all three-term harmonic progressions $a, b, c$ of strictly increasing positive integers in which $a=20$ and $b$ divides $c$.


Solution: Since $20, b, c$ are in harmonic progression, we have

$$
\frac{1}{20}+\frac{1}{c}=\frac{2}{b}
$$

which reduces to $b c+20 b-40 c=0$. This may also be written in the form

$$
(40-b)(c+20)=800
$$

Thus we must have $20<b<40$ or, equivalently, $0<40-b<20$. Let us consider the factorisation of 800 in which one term is less than 20 :

$$
\begin{aligned}
(40-b)(c+20)=800=1 \times 800=2 & \times 400=4 \times 200 \\
& =5 \times 160=8 \times 100=10 \times 80=16 \times 50
\end{aligned}
$$

We thus get the pairs

$$
(b, c)=(39,780),(38,380),(36,180),(35,140),(32,80),(30,60),(24,30)
$$

Among these 7 pairs, we see that only 5 pairs $(39,780),(38,380),(36,180),(35,140)$, $(30,60)$ fulfill the condition of divisibility: $b$ divides $c$. Thus there are 5 triples satisfying the requirement of the problem.
