Question

1. immediately following the ap precalculus exam, memes are created and shared online. the number of people that have seen the meme can be modeled using a geometric sequence, where one hour after the exam ends is hour 1. at hour 3, a total of 2500 people had seen the meme, and 20,000 people had seen the meme by hour 6. using this model, how many people will have seen the meme by hour 10?

Explanation:

Step1: Identify the geometric - sequence formula

The formula for the $n$th term of a geometric sequence is $a_n=a_1r^{n - 1}$, and the sum of the first $n$ terms of a geometric sequence is $S_n=\frac{a_1(1 - r^n)}{1 - r}$ when $r
eq1$. Here, $a_1$ is the number of people who saw the meme at hour 1, $r$ is the common ratio, and $n$ is the number of terms. We know that at hour 1 ($n = 1$), assume $a_1$ is some value, at hour 6 ($n = 6$), $S_6=20000$ and at hour 3 ($n = 3$), $S_3 = 2500$.

Step2: Use the sum - of - geometric - sequence formula

The sum formula for a geometric sequence $S_n=\frac{a_1(1 - r^n)}{1 - r}$. For $n = 3$, we have $S_3=\frac{a_1(1 - r^3)}{1 - r}=2500$. For $n = 6$, we have $S_6=\frac{a_1(1 - r^6)}{1 - r}=20000$.
Since $1 - r^6=(1 - r^3)(1 + r^3)$, we can write $\frac{S_6}{S_3}=\frac{\frac{a_1(1 - r^6)}{1 - r}}{\frac{a_1(1 - r^3)}{1 - r}}=\frac{1 - r^6}{1 - r^3}$. Substituting the values of $S_3$ and $S_6$, we get $\frac{20000}{2500}=\frac{1 - r^6}{1 - r^3}$. Since $\frac{1 - r^6}{1 - r^3}=1 + r^3$, then $1 + r^3 = 8$, so $r^3=7$, and $r=\sqrt[3]{7}$.
Substitute $r$ into the equation $S_3=\frac{a_1(1 - r^3)}{1 - r}=2500$. Since $1 - r^3=1 - 7=-6$, we have $\frac{a_1\times(- 6)}{1 - r}=2500$.

Step3: Find the sum at $n = 10$

We use the formula $S_{10}=\frac{a_1(1 - r^{10})}{1 - r}$. First, from $S_3=\frac{a_1(1 - r^3)}{1 - r}=2500$ and $r^3 = 7$, we can find $a_1$. After finding $a_1$, we calculate $S_{10}=\frac{a_1(1 - r^{10})}{1 - r}$. Another way is to note that $S_n=\frac{a_1(1 - r^n)}{1 - r}$, and we know the relationship between the sums.
We know that $S_n$ is a geometric - series sum. We can also use the property of geometric series. The sum of a geometric series $S_n$ with first term $a_1$ and common ratio $r$:
We know that $S_n=\frac{a_1(1 - r^n)}{1 - r}$. From $S_3 = 2500=\frac{a_1(1 - r^3)}{1 - r}$ and $S_6 = 20000=\frac{a_1(1 - r^6)}{1 - r}$.
Since $S_{10}=\frac{a_1(1 - r^{10})}{1 - r}$, and from $S_3=\frac{a_1(1 - r^3)}{1 - r}=2500$, we know $a_1=\frac{2500(1 - r)}{1 - r^3}$.
Substitute into $S_{10}$ formula:
First, since $r^3 = 7$, we know that $S_{10}=\frac{\frac{2500(1 - r)}{1 - r^3}(1 - r^{10})}{1 - r}=\frac{2500(1 - r^{10})}{1 - r^3}$.
$r^{10}=(r^3)^{\frac{10}{3}}=7^{\frac{10}{3}}$.
$S_{10}=\frac{2500(1 - 7^{\frac{10}{3}})}{1 - 7}$.
$S_{10}=\frac{2500(7^{\frac{10}{3}}-1)}{6}\approx\frac{2500(259.60 - 1)}{6}=\frac{2500\times258.60}{6}=107750$.

Answer:

107750