$\because x_{1}$，$x_{2}$，$x_{3}$，$\ldots$，$x_{10}$的平均数为$a$

$\therefore x_{1}+x_{2}+x_{3}+\ldots +x_{10}=10a$

$\therefore x_{1}+1$，$x_{2}+2$，$\ldots$，$x_{10}+10$的平均数$=\left(10a+1+2+3\ldots +9+10\right)\div 10=a+5.5$.

故答案为$a+5.5$.

$\because x_{1}$，$x_{2}$，$x_{3}$，$\ldots$，$x_{10}$的平均数为$a$

$\therefore x_{1}+x_{2}+x_{3}+\ldots +x_{10}=10a$

$\therefore x_{1}+1$，$x_{2}+2$，$\ldots$，$x_{10}+10$的平均数$=\left(10a+1+2+3\ldots +9+10\right)\div 10=a+5.5$.

故答案为$a+5.5$.