Given system of equation is,
$$
\begin{aligned}
& x+2 y+z=1 \\
& x+5 y+10 z=k^2 \\
\therefore & \quad D=\left|\begin{array}{lll}
1 & 2 & 1 \\
1 & 3 & 4 \\
1 & 5 & 10
\end{array}\right| \\
= & 1(30-20)-2(10-4)+1(5-3)=10-12+2=0
\end{aligned}
$$
Since, $\quad D=0$
$\therefore$ Given system of equation is consistent.
Therefore, $D_1=0$
$$
\begin{aligned}
& D_1=\left|\begin{array}{ccc}
1 & 2 & 1 \\
k & 3 & 4 \\
k^2 & 5 & 10
\end{array}\right| \\
& \Rightarrow 1(30-20)-2\left(10 k-4 k^2\right)+\left(5 k-3 k^2\right)=0 \\
& \Rightarrow \quad 10-20 k+8 k^2+5 k-3 k^2=0 \\
& \Rightarrow \quad 5 k^2-15 k+10=0 \\
& \Rightarrow \quad k^2-3 k+2=0 \\
& \Rightarrow \quad(k-2)(k-1)=0 \\
& \Rightarrow \quad k=2,1 \\
&
\end{aligned}
$$
Hence, the real values of $k$ i.e.
$$
\begin{array}{rlrl}
A & =2 \text { and } B=1 \\
\therefore & & A+B & =2+1=3
\end{array}
$$