No.

Consider the sequence $3, 1011, \dots$, where the first two terms have sum of digits 3. The difference 1008 is divisible by 16, so all terms must have remainder 3 modulo 16. We show that a) among those numbers $16k+3$, the minimum sum of last 4 digits is 3, and b) there are only two such numbers up to congruence modulo 10000 (i.e. $0003$ and $1011$). Given that such number is odd, if it's sum of last 4 digits is at most 3, then it must end with 1 or 3. In the second case the last 4 digits can only be $0003$. In the first case, since it's congruent to 3 mod 4, the tens digits must be odd, so the last two digits can only be $11$. Now that the 100s and 1000s digits have to be $\{0, 1\}$ (while cannot both be 1), we only need to consider $0011, 0111, 1011$, and only $1011$ has remainder 3 modulo 16. This justifies our claim. Circling back to the problem, we see that all such numbers in our arithmetic progression with sum of digits = 3 must have the form $0003$ or $1011$ in the last 4 digits, and 0 elsewhere, which means there can only be two of them.

pick $3,1011...$ then see their modular form and rule out the possibilities