Let $a_1,$ $a_2,$ $a_3,$ $\dots$ be a sequence. If
\[a_n = a_{n - 1} + a_{n - 2}\]
for all $n \ge 3,$ and $a_{11} = 1$ and $a_{10} = 4,$ then find $a_6.$
Working backwards,
\(a_9 = 1 - 4 = -3 \)
\(a_8 = 4 - (-3) = 7\)
\(a_7 = -3-7=-10\)
\(a_6 = 7-(-10)=\boxed{17}\)