Problem 3
Suppose that is a strictly increasing sequence of positive integers such that the subsequences and are both arithmetic progressions. Prove that the sequence is itself an arithmetic progression.
Step 5 of 5: The increment is forced to be constant
In plain words
Once the largest and smallest increments coincide, every increment in between must equal that common value too.
Detailed analysis
Steps 3 and 4 give . Since for every and , every equals this common value. So is constant, meaning is an arithmetic progression, as required.