Solution:

(i) $4,2,1,1 / 2 \ldots$ to 10 terms
It is known that, the sum of GP for $n$ terms $=a\left(r^{n}-1\right) /(r-1)$
Given that,
$\mathrm{a}=4, \mathrm{r}=\mathrm{t}_{2} / \mathrm{t}_{1}=2 / 4=1 / 2, \mathrm{n}=10$
Substitute the values in
$\begin{array}{l}
\mathrm{a}\left(\mathrm{r}^{\mathrm{n}}-1\right) /(\mathrm{r}-1)=4\left((1 / 2)^{10}-1\right) /((1 / 2)-1) \\
=4\left((1 / 2)^{10}-1\right) /((1-2) / 2) \\
=4\left((1 / 2)^{10}-1\right) /(-1 / 2)
\end{array}$
$\begin{array}{l}
=4\left((1 / 2)^{10}-1\right) \times-2 / 1 \\
=-8[1 / 1024-1] \\
=-8[1-1024] / 1024 \\
=-8[-1023] / 1024 \\
=1023 / 128
\end{array}$