Solution:
(i) $2 / 9-1 / 3+1 / 2-3 / 4+\ldots$ to 5 terms;
Given that
$\begin{array}{l}
a=2 / 9 \\
r=t_{2} / t_{1}=(-1 / 3) /(2 / 9)=-3 / 2 \\
n=5
\end{array}$
Using the formula,
The sum of GP for $n$ terms $=a\left(1-r^{n}\right) /(1-r)$
$\begin{array}{l}
\mathrm{a}\left(1-\mathrm{r}^{\mathrm{n}}\right) /(1-\mathrm{r})=(2 / 9)\left(1-(-3 / 2)^{5}\right) /(1-(-3 / 2)) \\
=(2 / 9)\left(1+(3 / 2)^{5}\right) /(1+3 / 2) \\
=(2 / 9)\left(1+(3 / 2)^{5}\right) /(5 / 2) \\
=(2 / 9)(1+243 / 32) /(5 / 2) \\
=(2 / 9)((32+243) / 32) /(5 / 2) \\
=(2 / 9)(275 / 32) \times 2 / 5 \\
=55 / 72
\end{array}$
(ii) $(x+y)+\left(x^{2}+x y+y^{2}\right)+\left(x^{3}+x^{2} y+x y^{2}+y^{3}\right)+\ldots$ to $n$ terms;
Let $S_{n}=(x+y)+\left(x^{2}+x y+y^{2}\right)+\left(x^{3}+x^{2} y+x y^{2}+y^{3}\right)+\ldots .$ to $n$ terms
Multiply and divide by $(x-y)$ we obtain,
$\mathrm{S}_{n}=1 /(\mathrm{x}-\mathrm{y})\left[(\mathrm{x}+\mathrm{y})(\mathrm{x}-\mathrm{y})+\left(\mathrm{x}^{2}+\mathrm{xy}+\mathrm{y}^{2}\right)(\mathrm{x}-\mathrm{y}) \ldots\right.$ upto $\mathrm{n}$ terms $]$ $(x-y) S_{n}=\left(x^{2}-y^{2}\right)+x^{3}+x^{2} y+x y^{2}-x^{2} y-x y^{2}-y^{3}$..upto $n$ terms $(x-y) S_{n}=\left(x^{2}+x^{3}+x^{4}+\ldots n\right.$ terms $)-\left(y^{2}+y^{3}+y^{4}+\ldots n\right.$ terms $)$
Using the formula,
The sum of GP for $n$ terms $=a\left(1-r^{n}\right) /(1-r)$
In the above sum we have two G.Ps, so,
$\begin{array}{l}
(x-y) S_{n}=x^{2}\left[\left(x^{n}-1\right) /(x-1)\right]-y^{2}\left[\left(y^{n}-1\right) /(y-1)\right] \\
S_{n}=1 /(x-y)\left\{x^{2}\left[\left(x^{n}-1\right) /(x-1)\right]-y^{2}\left[\left(y^{n}-1\right) /(y-1)\right]\right\}
\end{array}$