Solution:
(i) $5+55+555+\ldots$ to $n$ terms.
Let’s take 5 as a common term therefore we obtain,
$5[1+11+111+\ldots \mathrm{n}$ terms $]$
Now multiplying and dividing by 9 we obtain,
$5 / 9[9+99+999+\ldots \mathrm{n}$ terms $]$
$5 / 9\left[(10-1)+\left(10^{2}-1\right)+\left(10^{3}-1\right)+\ldots \mathrm{n}\right.$ terms $]$
$5 / 9\left[\left(10+10^{2}+10^{3}+\ldots \mathrm{n} \text { terms }\right)-\mathrm{n}\right]$
Therefore the G.P is
$5 / 9\left[\left(10+10^{2}+10^{3}+\ldots \mathrm{n} \text { terms }\right)-\mathrm{n}\right]$
Using the formula,
The sum of GP for $n$ terms $=\mathrm{a}\left(r^{n}-1\right) /(r-1)$
Where, $a=10, r=10^{2} / 10=10, n=n$
$\begin{array}{l}
a\left(r^{n}-1\right) /(r-1)= \\
\quad=\frac{5}{9}\left\{10 \times \frac{\left(10^{n}-1\right)}{10-1}-n\right\} \\
\quad=\frac{5}{9}\left\{\frac{10}{9}\left(10^{n}-1\right)-n\right\} \\
=\frac{5}{81}\left\{10^{n+1}-9 n-10\right\}
\end{array}$
(ii) $7+77+777+\ldots$ to $\mathrm{n}$ terms.
Let’s take 7 as a common term therefore we obtain,
Now multiplying and dividing by 9 we obtain,
$7 / 9[9+99+999+\ldots n$ terms $]$
$7 / 9\left[(10-1)+\left(10^{2}-1\right)+\left(10^{3}-1\right)+\ldots+\left(10^{n}-1\right)\right]$ $7 / 9\left[\left(10+10^{2}+10^{3}+\ldots+10^{n}\right)\right]-7 / 9[(1+1+1+\ldots$ to $n$ terms $)]$
Therefore the terms are in G.P
Where, $a=10, r=10^{2} / 10=10, n=n$
Using the formula,
The sum of GP for $n$ terms $=a\left(r^{n}-1\right) /(r-1)$
$\begin{array}{l}
7 / 9\left[10\left(10^{n}-1\right) /(10-1)\right]-n \\
7 / 9\left[10 / 9\left(10^{n}-1\right)-n\right] \\
7 / 81\left[10\left(10^{n}-1\right)-n\right] \\
7 / 81\left(10^{n+1}-9 n-10\right)
\end{array}$