dazedandconfused wrote:
What I am confused by is the fact that if we use the equation of a geometric series, we get the value of the tenth term to be \(512*(\text{the original population})\). This is because we know that \(a_n=a_1*r^{n-1}\). Using this formula, given that r=2, \(a_n=a_1*2^{10-1}=512*a_1\).
How is it that we cannot use the geometric series to find the 10th term in this setting?
If \(a_1\) represents the INITIAL population, then \(a_2\) represents the population after 1 "doubling"
So, \(a_2=a_1*2^{1}=a_1*(2)\)
Likewise, \(a_3\) represents the population after 2 "doublings"
So, \(a_3=a_1*2^{2}=a_1*(4)\)
Likewise, \(a_4\) represents the population after 3 "doublings"
So, \(a_4=a_1*2^{3}=a_1*(8)\)
In general, \(a_n\) represents the population after n-1 "doublings"
So, \(a_n=a_1*r^{n-1}\)
30/3 = 10
So, the population gets doubled 10 times.
Since \(a_n\) represents the population after n-1 "doublings," we need to find the value of \(a_{11}\)
We get: \(a_{11}=a_1*2^{10}=a_1*1024\)
Cheers,
Brent
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Brent Hanneson - founder of Greenlight Test Prep