$a+a r+a r^{2}+\ldots+a r^{n-1}=a\left(\frac{r^{n}-1}{r-1}\right), r \neq 1$
Let P(n) be the given statement.
Now,
Step 1;
$P(n)=a+a r+a r^{2}+\ldots+a r^{n-1}=a\left(\frac{r^{n}-1}{r-1}\right), r \neq 1$
$P(1)=a=a\left(\frac{r^{1}-1}{r-1}\right)$
Hence, $P(1)$ is true.
Step 2 :
Suppose $P(m)$ is true.
Then,
$a+a r+a r^{2}+\ldots+a r^{m-1}=a\left(\frac{r^{m}-1}{r-1}\right), r \neq 1$
To show: $P(m+1)$ is true whenever $P(m)$ is true.
That is,
$a+a r+a r^{2}+\ldots+a r^{m}=a\left(\frac{r^{m+1}-1}{r-1}\right), r \neq 1$
$W e$ know that $P(m)$ is true.
Thus, we have :
$a+a r+a r^{2}+\ldots+a r^{m-1}=a\left(\frac{r^{m}-1}{r-1}\right)$
$\Rightarrow a+a r+a r^{2}+\ldots+a r^{m-1}+a r^{m}=a\left(\frac{r^{m}-1}{r-1}\right)+a r^{m} \quad$ [Adding $a r^{m}$ to both sides]
$\Rightarrow P(m+1)=a\left(\frac{r^{m}-1+r \cdot r^{m}-r^{m}}{r-1}\right)$
$\Rightarrow P(m+1)=a\left(\frac{r^{m+1}-1}{r-1}\right), r \neq 1$
Thus, $P(m+1)$ is true.
By the principle of $m$ athematical $i$ nduction, $P(n)$ is true for all $n \in N$.
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