Place the digits 1 to 9, using each digit only once so that:

- the 1 and the 2 and all the numbers between them add up to 12
- the 2 and the 3 and all the numbers between them add up to 23
- the 3 and the 4 and all the numbers between them add up to 34
- the 4 and the 5 and all the numbers between them add up to 45

472918365

Main given: "4 and 5..."; right, MxM?

Since 1 to 9 sums to 45:
1st: 4 or 5
2nd to 8th: other 7 digits
9th: 5 or 4

A perfect solve, manxman!

4 and 5 must be at the ends since the sum of 1 through 9 is 45

4.......5

The sum of all numbers between and including 4 and 3 is 34. Add the 5 at the other end for 39 and that leaves 6 (45-39). It must BE 6, not 1 and 5 or 2 and 4, because the 4 and 5 are already claimed

4.....365

And keep going.