This is a challenging problem from the c.100AD Chinese mathematical work, Jiǔ zhāng suàn shù (The Nine Chapters on the Mathematical Art) found at the MAA Convergence website.
“Now a good horse and an inferior horse set out from Chang’an to Qi. Qi is 3000 li from Chang’an. The good horse travels 193 li on the first day and daily increases by 13 li; the inferior horse travels 97 li on the first day and daily decreases by ½ li. The good horse reaches Qi first and turns back to meet the inferior horse. Tell: how many days until they meet and how far has each traveled?”
The solution involves common fractions, which the Chinese were already adept at using by 100 BC.
See Horses to Qi for a solution.
The solution can be shortened with following assumptions.
Let the strong horse start from point A and the weak horse start from point B towards each other and meet on the n’th day. The initial distance between them is 6000 Li.
Since they are approaching each other there speeds will be added.
The first term a=193+97=290
Common difference = 13 + (-0.5) = 12.5
Hence the sum of distance travelled by both together in n days will be 6000 Li.
(n/2)*[2*290 + (n-1)*12.5] = 6000
Solving, n = 15.70950 days.
Rest follows.