$u_{10}=u_1.q^9$
$\Rightarrow \dfrac{1}{1024}=\dfrac{1}{2}.q^9$
$\Leftrightarrow q^9=\dfrac{1}{512}$
$\Leftrightarrow q=\sqrt[9]{\dfrac{1}{512}}=\dfrac{1}{2}$
$\to u_{20}=u_1.q^{19}=\dfrac{1}{2^{20}}$
Đặt $u_n=\dfrac{1}{4096}$
$u_n=u_1.q^{n-1}$
$\Rightarrow \dfrac{1}{4096}=\dfrac{1}{2}.\dfrac{1}{2^{n-1}}$
$\Rightarrow n=12$
$\to \dfrac{1}{4096}$ là số hạng thứ $12$.