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Re: A cockroach population doubles every 3 days. In 30 days, by [#permalink]
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sandy wrote:
Explanation

The percent increase is the difference between the amounts divided by the original, converted to a percent. If the population doubles, mathematically the increase can be written as a
power of 2. In the 30-day interval, if the original population is 1, it will double to 2 after three days —so, 21 represents the population after the first increase, the second increase would then be 22 and so on. Since there are 10 increases, the final population would be 210 or 1,024. Therefore, the difference, 1,024 – 1, is 1,023. Use the percent change formula to calculate percent increase:

Percent Change=\((\frac{Difference}{original} \times 100)\)
Percent Change=\((\frac{1023}{1} \times 100)=102300 \%\)

Note that the new number is 102,400% of the original, but that was not the question asked—the percent increase is 102,300%.



FROM 2ND LINE 21 represents the population after the first increase, the second increase would then be 22
This will be: \(2^1\) represents the population after the first increase, the second increase would then be \(2^2\)
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Re: A cockroach population doubles every 3 days. In 30 days, by [#permalink]
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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?
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Re: A cockroach population doubles every 3 days. In 30 days, by [#permalink]
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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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Re: A cockroach population doubles every 3 days. In 30 days, by [#permalink]
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Hi Brent,

Just re-read your response. This was very very helpful. Thanks so much!
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Re: A cockroach population doubles every 3 days. In 30 days, by [#permalink]
Hello,

I understand this for the most part, but why are we subtracting 1 from the answer? It can make the difference between a right answer and a wrong one. why is it D and not E?
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Re: A cockroach population doubles every 3 days. In 30 days, by [#permalink]
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Re: A cockroach population doubles every 3 days. In 30 days, by [#permalink]
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