Question

a bacteria culture begins with 11 bacteria that triple in amount at the end of every hour. how many bacteria are grown during the 7th hour?
1,331
8,019
9,317
24,057

Explanation:

Step1: Find the formula for the number of bacteria

The number of bacteria forms a geometric - sequence with the first term $a_1 = 11$ and the common ratio $r = 3$. The formula for the $n$th term of a geometric sequence is $a_n=a_1r^{n - 1}$. So, the number of bacteria at the end of the $n$th hour is $a_n = 11\times3^{n - 1}$.

Step2: Find the number of bacteria at the end of the 6th hour

Substitute $n = 6$ into the formula $a_n=11\times3^{n - 1}$. We get $a_6=11\times3^{6 - 1}=11\times3^{5}=11\times243 = 2673$.

Step3: Find the number of bacteria at the end of the 7th hour

Substitute $n = 7$ into the formula $a_n=11\times3^{n - 1}$. We get $a_7=11\times3^{7 - 1}=11\times3^{6}=11\times729 = 8019$.

Step4: Calculate the number of bacteria grown in the 7th hour

The number of bacteria grown in the 7th hour is the number of bacteria at the end of the 7th hour minus the number of bacteria at the end of the 6th hour. So, $a_7 - a_6=8019-2673 = 5346$.

However, if we consider the growth from the start of the 7th hour to the end of the 7th hour, we can also calculate it as follows:
The number of bacteria at the start of the 7th hour is $a_6 = 11\times3^{5}$ and at the end is $a_7 = 11\times3^{6}$. The growth is $11\times3^{6}-11\times3^{5}=11\times3^{5}(3 - 1)=11\times243\times2=5346$.

If we assume the problem means the number of bacteria at the end of the 7th hour (a mis - understanding of the problem statement perhaps), then:

Step1: Use the geometric - sequence formula

The formula for the number of bacteria at the end of the $n$th hour is $a_n = 11\times3^{n - 1}$.

Step2: Substitute $n = 7$

$a_7=11\times3^{7 - 1}=11\times3^{6}=11\times729 = 8019$.

Answer:

8,019