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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 $?$
$120,480,30(2)^{n}$
$120,480,30(2)^{n}$
$120,480,30(2)^{n}$
$120,480,30(2)^{n}$
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 $r=2 \quad \therefore a_{3}=a r^{2}=(30)(2)^{2}=120$
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 $30(2)^{n}$
