Let $P(n): a+a r+a r^2+\ldots+a r^{n-1}$
$=\frac{a\left(1-r^n\right)}{1-r}, r \neq 1 \quad \ldots(i)$
For $n=1$, L.H.S $=a$ and R.H.S. $=\frac{a(1-r)}{1-r}=a$
$\therefore \quad$ L.H.S. $=$ R.H.S. i.e., $P(\mathrm{i})$ is true.
Let us suppose that $\mathrm{P}(k)$ is true.
Putting $n=k$ in (i), we have,
$a+a r+a r^2+\ldots a r^{k-1}=\frac{a\left(1-r^k\right)}{1-r}+a r^k \quad \ldots(ii)$
Changing $k$ to $k+1$ in the last term $a r^{k-1}$ or L.H.S. of
(ii), it becomes $a r^{k+1-1}$ i.e., $a r^k$.
Adding $a r^k$ to both sides of (ii), we have $a+a r+a r^2+\ldots+a r^{k-1}+a r^k$
$\begin{aligned}
&=\frac{a\left(1-r^k\right)}{1-r}+a r^k \\
&=\frac{a\left(1-r^k\right)+a r^k(1-r)}{1-r}=\frac{a-a r^k+a r^k-a r^{k+1}}{1-r} \\
&=\frac{a-a r^{k+1}}{1-r}=\frac{a\left(1-r^{k-1}\right)}{1-r} \quad \ldots(iii)
\end{aligned}$
$\therefore P(n)$ is true for $n=k+1$ i.e., $P(k+1)$ is true.
$\therefore$ By Principle of Mathematical Induction, $\mathrm{P}(n)$ is true for natural numbers $n$.