Ta có: $\left \{ {{u_1+u_2=6} \atop {u_{10}-u_2}=8} \right.$ ⇔ $\left \{ {{2u_1+4d=6} \atop {8d=8}} \right.$ ⇔ $\left \{ {{u_1=1} \atop {d=1}} \right.$
$A=u_1.u_2+u_2.u_3+...+u_{2019}.u_{2020}$
$=1.2+2.3+...+2019.2020$
⇒ $3A=1.2.3+2.3.3+...+2019.2020.3$
$=1.2.(3-0)+2.3.(4-1)+...+2019.2020.(2021-2018)$
$=(1.2.3+2.3.4+...+2019.2020.2021)-(0.1.2+1.2.3+...+2018.2019.2020)$
$=2019.2020.2021$
⇒ $A=\frac{2019.2020.2021}{3}$
Vậy $S=$$\frac{2020.2021}{3}=$ $\frac{4082420}{3}$