Answer
$\sum\limits_{i =1}^{n}ar^{i-1}=a+ar+ar^{2}+....+ar^{n-1}=\frac{a(r^{n}-1)}{r-1}$
Work Step by Step
Consider $\sum\limits_{i =1}^{n}ar^{i-1}=a+ar+ar^{2}+....+ar^{n-1}$
$r\times\sum\limits_{i =1}^{n}ar^{i-1}=r\times (a+ar+ar^{2}+....+ar^{n-1})$
$r\times\sum\limits_{i =1}^{n}ar^{i-1}=ar+ar^{2}+....+ar^{n-1}+ar^{n}$
$r\sum\limits_{i =1}^{n}ar^{i-1}-\sum\limits_{i =1}^{n}ar^{i-1}=ar^{n}-a$
$(r-1)\sum\limits_{i =1}^{n}ar^{i-1}=a(r^{n}-1)$
$\sum\limits_{i =1}^{n}ar^{i-1}=\frac{a(r^{n}-1)}{r-1}$
Hence, $\sum\limits_{i =1}^{n}ar^{i-1}=a+ar+ar^{2}+....+ar^{n-1}=\frac{a(r^{n}-1)}{r-1}$