Fly and cyclist

Two cyclists are 20 km apart on the same line. They start at the same time as each other at a speed of 10 km/hr. A fly sitting on one of the cyclist's handles starts flying toward the other cyclists at a speed of 20 km/hr. It touches the handle and moves back to the first one. Likewise, it goes on till the two cyclists meet each other. What is the distance the fly traveled, and how many times it moved to and fro?

Final Answer:

x =  20 km
n = INF

Step-by-step explanation:

$$\begin{gathered} v_1=10\text{km/h} \\ {v}_2=20\text{km/h} \\ s=20\text{km} \\ {v}_2>v_1 \\ {t}_1=s/(v_1+v_2)=20/(10+20)=\frac{2}{3}\doteq 0.6667\text{h} \\ {s}_1=v_2\cdot t_1=20\cdot \frac{2}{3}=\frac{20\cdot 2}{3}=\frac{40}{3}\doteq 13.3333\text{km} \\ {t}_2=(s-2\cdot t_1\cdot v_1)/(v_1+v_2)=(20-2\cdot \frac{2}{3}\cdot 10)/(10+20)=\frac{2}{9}\doteq 0.2222\text{h} \\ {s}_2=v_2\cdot t_2=20\cdot \frac{2}{9}=\frac{20\cdot 2}{9}=\frac{40}{9}\doteq 4.4444\text{km} \\ {t}_3=(s-2\cdot (t_1+t_2)\cdot v_1)/(v_1+v_2)=(20-2\cdot (\frac{2}{3}+\frac{2}{9})\cdot 10)/(10+20)=\frac{2}{27}\doteq 0.0741\text{h} \\ {s}_3=v_2\cdot t_3=20\cdot \frac{2}{27}=\frac{20\cdot 2}{27}=\frac{40}{27}\doteq 1.4815\text{km} \\ {q}_1=t_2/t_1=\frac{2}{9}/\frac{2}{3}=\frac{2}{9}:\frac{2}{3}=\frac{2}{9}\cdot \frac{3}{2}=\frac{2\cdot 3}{9\cdot 2}=\frac{6}{18}=\frac{1}{3}\doteq 0.3333 \\ {q}_2=t_3/t_2=\frac{2}{27}/\frac{2}{9}=\frac{2}{27}:\frac{2}{9}=\frac{2}{27}\cdot \frac{9}{2}=\frac{2\cdot 9}{27\cdot 2}=\frac{18}{54}=\frac{1}{3}\doteq 0.3333 \\ q=q_1=\frac{1}{3}\doteq 0.3333 \\ {a}_1=t_1=\frac{2}{3}\doteq 0.6667 \\ t=t_1+t_2+t_3+\dots \\ t=\frac{a_1}{1-q}=\frac{0.6667}{1-0.3333}=1\text{h} \\ x=v_2\cdot t=20\cdot 1=20\text{km} \end{gathered}$$

$n=\infty$

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