The possible scores that can be obtained are: $1,2,3,4,5,6,8,9,10,12,15,16,18,20,24,25,30,36$.
The third and fourth scores differ by $11$. The only pairs of numbers that do this are $(1,12)$, $(4,15)$, $(5,16)$, $(9,20)$ and $(25,36)$.
When these pairs are completed into sequences, they become:
Of these, only the fourth one uses only accessible numbers, so the sequence is $10,15,9,20,12$.
This problem is taken from the UKMT Mathematical Challenges.