若等差数列 $\left\{a_{n}\right\}$ 满足 $a_{1}=\frac{1}{2}, a_{4}+a_{6}=5$, 则公差 $d=$ $a_{2}+a_{4}+a_{6}+\cdots+a_{20}=$ \$ \qquad \$ .
等差数列 $\left\{a_{n}\right\}$ 满足 $\mathrm{a}_{1}=\frac{1}{2}, \mathrm{a}_{4}+\mathrm{a}_{6}=5=2 \mathrm{a}_{5}, \therefore \mathrm{a}_{5}=\frac{5}{2}$,

$\therefore \frac{5}{2}=\frac{1}{2}+4 \mathrm{~d}$, 则公差 $\mathrm{d}=\frac{1}{2}$.

$\therefore \mathrm{a}_{2}+\mathrm{a}_{4}+\mathrm{a}_{6}+\ldots+\mathrm{a}_{20}=10\left(\mathrm{a}_{1}+\mathrm{d}\right)+\frac{10 \times 9}{2} \times 2 \mathrm{~d}=10 \times 1+45=55$,

故答案为: $\frac{1}{2}, 55$.