The number of bacteria in a certain culture d | vedclass

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(A) The number of bacteria doubles every hour,forming a geometric progression $(G.P.)$ where the initial number of bacteria $a = 30$ and the common ra

Vedclass · Accepted Answer

$120, 480, 30(2)^{n}$

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(A) The number of bacteria doubles every hour,forming a geometric progression $(G.P.)$ where the initial number of bacteria $a = 30$ and the common ratio $r = 2$.The number of bacteria after $t$ hours is given by $a_{t+1} = a 	imes r^{t} = 30 	imes 2^{t}$.At the end of the $2^{	ext{nd}}$ hour $(t=2)$: $30 	imes 2^{2} = 30 	imes 4 = 120$.At the end of the $4^{	ext{th}}$ hour $(t=4)$: $30 	imes 2^{4} = 30 	imes 16 = 480$.At the end of the $n^{	ext{th}}$ hour $(t=n)$: $30 	imes 2^{n}$.

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?

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