Let $F_n$ be the number of such qualified strings.

• If the $n^{th}$ letter is $B$, then there are  $F_{n-1}$ possibilities. 
• If the $n^{th}$ letter is $A$, then the $(n-1)^{th}$ letter must be $B$. Thus, there are $F_{n-2}$ possibilities.

It follows that $F_n$ satisfies the following recursion when $n \ge 3$: $$F_n=F_{n-1}+F_{n-2}$$

Meanwhile, it is easy to compute that $F_1=2$, and $F_2=2^2-1=3$. Hence,

• $F_3=2+3=5$
• $F_4=5+3=8$
• $F_5=8+5=13$
• $F_6=13+8=21$
• $F_7=21+13=34$
• $F_8=34+21=55$
• $F_9=55+34=89$
• $F_{10}=89+55=\boxed{144}$