10. 已知数列 $-1, a_{1}, a_{2},-4$ 成等差数列, $-1, b_{1}, b_{2}, b_{3},-4$ 成等比数列, 则 $\frac{a_{2}-a_{1}}{b_{2}}$的值是
because-1, a_{1}, a_{2},-4$ 成等差数列, 设公差为 $d$,

则 $a_{2}-a_{1}=d=\frac{1}{3}[(-4)-(-1)]=-1$,

$\because-1, b_{1}, b_{2}, b_{3},-4$ 成等比数列,

$\therefore b_{2}^{2}=(-1) \times(-4)=4, \quad \therefore b_{2}= \pm 2$.

若设公比为 $q$, 则 $b_{2}=(-1) q^{2}, \therefore b_{2}<0$.

$\therefore b_{2}=-2, \therefore \frac{a_{2}-a_{1}}{b_{2}}=\frac{-1}{-2}=\frac{1}{2}$.