Question:
The number of bacteria in a certain culture doubles every hour. If there were 30 bacteria present in the culture originally, how many bacteria will be present at the end of $2^{\text {nd }}$ hour, $4^{\text {th }}$ hour and $n^{\text {th }}$ hour?
Solution:
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 $r=2$
$\therefore a_{3}=a r^{2}=(30)(2)^{2}=120$
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 $n^{\text {th }}$ hour will be $30(2)^{n}$.