8. 已知 $\left\{a_{n}\right\}$ 为等差数列, $a_{1}+a_{3}+a_{5}=105, a_{2}+a_{4}+a_{6}=99$, 则 $a_{20}=$
$\because a_{1}+a_{3}+a_{5}=105, \therefore 3 a_{3}=105, a_{3}=35$.

$\therefore a_{2}+a_{4}+a_{6}=3 a_{4}=99$.

$\therefore a_{4}=33, \quad \therefore d=a_{4}-a_{3}=-2$.

$\therefore a_{20}=a_{4}+16 d=33+16 \times(-2)=1$.