Not every finite sequence can be obtained by a linear congruence generator. In fact, the one you request cannot.
Note that we must have $m\geq 23$ in order to "get" $22$ as an answer. We have that you are requiring: $\begin{align*} 2a + c &\equiv 11 \pmod{m}\\ 11a+c &\equiv 5\pmod{m}\\ 5a+c &\equiv 9\pmod{m}\\ 9a+c &\equiv 6\pmod{m}\\ 6a+c&\equiv 22\pmod{m}. \end{align*}$
Subtracting the first congruence from the second, we get $9a \equiv -6\pmod{m}$; subtracting this from the fourth congruence, we obtain $c\equiv 12\pmod{m}$. So now we know the value of $c$, which gives $\begin{align*} 2a &\equiv -1\pmod{m}\\ 11a &\equiv -7\pmod{m}\\ 5a &\equiv -3\pmod{m}\\ 9a &\equiv -6\pmod{m}\\ 6a &\equiv 10\pmod{m}. \end{align*}$ Multiply the first congruence by $3$ to get $6a\equiv -3\pmod{m}$. Since we also need $6a\equiv 10\pmod{m}$, we must have $10\equiv -3\pmod{m}$, or $13\equiv 0\pmod{m}$. But that means that $m=1$ or $m=13$, which contradicts our requirement that $m\geq 23$.
There are other contradictions in this system: multiply the third congruence by $2$ to get $10a\equiv -6\pmod{m}$, and subtracting from the second congruence we get $a\equiv 3\pmod{m}$. But subtracting $9a\equiv -6\pmod{m}$ from $10a\equiv -6\pmod{m}$ would give $a\equiv 0\pmod{m}$, so we would need $3\equiv 0\pmod{m}$, again a problem.
Or note that $2a\equiv -1\pmod{m}$ tells you that $\gcd(2,m)=1$, so that $6a\equiv 10\pmod{m}$ is equivalent to $3a\equiv 5\pmod{m}$, and multiplying by $3$ gives $9a\equiv 15\pmod{m}$; comparing with $9a\equiv -6\pmod{m}$ gives $21\equiv 0\pmod{m}$, so $m$ would have to divide $21$. Etc.