Tổng 9 số hạng đầu:
$S=u_1+u_2+u_3+u_4+...+u_9$
$=9u_1+d+2d+3d+...+8d$
$=9u_1+36d$
$200<S<220$
$\Leftrightarrow 200<9u_1+36d<220$
Ta có $u_2=u_1+d=12\Leftrightarrow d=12-u_1$
$\Rightarrow 200<9u_1+36(12-u_1)<220$
$\Leftrightarrow 200< -27u_1+432<220$
$\Leftrightarrow 7,85<u_1<8,6$
$\Rightarrow u_1=8$
$\Rightarrow d=4$
Vậy $u_n=8+4(n-1)=4n+4$