QUESTION 7
Easy
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 the \(2^{nd}\) hour, \(4^{th}\) hour and \(n^{th}\) hour?
SOLUTION
1
Form the GP from what is given
Initial number of bacteria = 30
Since population doubles every hour, the GP is
\(30,\,60,\,120,\,240,...\)
where \(a = 30\) and \(r = 2\)
Since population doubles every hour, the GP is
\(30,\,60,\,120,\,240,...\)
where \(a = 30\) and \(r = 2\)
2
Find number of bacteria at the end of \(2^{nd}\) hour
Using the GP formula, \(T_n = ar^n\) (since 30 is the initial population at hour 0)
\(\begin{aligned} T_2 &= 30(2)^2 \\ &= 30 × 4 \\ &= 120\end{aligned}\)
At the end of the \(2^{nd}\) hour, there will be 120 bacteria.
\(\begin{aligned} T_2 &= 30(2)^2 \\ &= 30 × 4 \\ &= 120\end{aligned}\)
At the end of the \(2^{nd}\) hour, there will be 120 bacteria.
3
Find number of bacteria at the end of \(4^{th}\) hour
\(\begin{aligned} T_4 &= 30(2)^4 \\ &= 30 × 16 \\ &= 480\end{aligned}\)
At the end of the \(4^{th}\) hour, there will be 480 bacteria.
At the end of the \(4^{th}\) hour, there will be 480 bacteria.
4
Find number of bacteria at the end of \(n^{th}\) hour
\(T_n = 30(2)^n\)
At the end of the \(n^{th}\) hour, there will be \(30(2)^n\) bacteria.
At the end of the \(n^{th}\) hour, there will be \(30(2)^n\) bacteria.
Concept Note
When a quantity repeatedly doubles, triples, halves, etc., it forms a Geometric Progression.
If counting begins from Stage/Hour 0, the general formula is
\(T_n = ar^n\)