It is given that the number of bacteria doubles every hour. Therefore, the number of bacteria after every hour will form a $G.P.$

Here, $a=30$ and $r=2 \quad \therefore a_{3}=a r^{2}=(30)(2)^{2}=120$

Therefore, the number of bacteria at the end of $2^{\text {nd }}$ hour will be $120 .$

$a_{5}=a r^{4}=(30)(2)^{4}=480$

The number of bacteria at the end of $4^{\text {th }}$ hour will be $480 . $

$a_{n+1}=a r^{n}=(30) 2^{n}$

Thus, number of bacteria at the end of $n^{t h}$ hour will be $30(2)^{n}$