MathLabs

Problem 3

Suppose that s1,s2,s3,…s_1, s_2, s_3, \dots is a strictly increasing sequence of positive integers such that the subsequences ss1,ss2,ss3,…s_{s_1}, s_{s_2}, s_{s_3}, \dots and ss1+1,ss2+1,ss3+1,…s_{s_1+1}, s_{s_2+1}, s_{s_3+1}, \dots are both arithmetic progressions. Prove that the sequence s1,s2,s3,…s_1, s_2, s_3, \dots is itself an arithmetic progression.
Step 4 of 5: The symmetric argument pins the minimum too
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da′=m  ⟹  ds(s(a′))=⋯=ds(s(a′+1))−1=m  ⟹  m=B−Ad_{a'}=m \implies d_{s(s(a'))}=\dots=d_{s(s(a'+1))-1}=m \implies m=B-A
Detailed analysis

Repeat Step 3 verbatim at an index a′a' with da′=md_{a'}=m: the same telescoping identity DMDM becomes Dm=D(s(a′+1)−s(a′))Dm=D\big(s(a'+1)-s(a')\big), written as a sum of DD terms each ≥m\ge m by definition of mm, so their sum is ≥Dm\ge Dm, forcing every term to equal mm exactly. In particular ds(s(a′))=md_{s(s(a'))}=m, and Step 2 again gives ds(s(a′))=B−Ad_{s(s(a'))}=B-A. Hence m=B−Am=B-A as well.