Question

1. On the first day of a flu epidemic at your school, 20 students were absent. Each day forthe following week another 25% of the student population was absent.Part A: Determine how many students were absent for the first week of the flu epidemic.Complete the table below,DayStudentsAbsent1202345Part B: Determine the explicit formula to determine the number of students absent if theflu epidemic continues for more than one week,Explicit Formula:Part C: Determine the recursive formula to determine the number of students absent if theflu epidemic continues for more than one week,Sign ou

Answer 1

Given:

$\begin{gathered} a_3=2344 \\ a_{13}=8.029 \end{gathered}$

The recursive formula for the n th term is,

$a_{n-1}=a_0r^{n-1}\text{ ------(1)}$

Here, a0 is the first term and r is the common ratio.

Now, for a3,

$\begin{gathered} n-1=3 \\ n=4 \end{gathered}$

For a13,

$\begin{gathered} n-1=13 \\ n=14 \end{gathered}$

Now, we can write

$\begin{gathered} \frac{a_{13}}{a_3}=\frac{8.029}{2344} \\ \frac{a_0r^{13}}{a_0r^3}=\frac{8.029}{2344} \\ \frac{r^{13}}{r^3}=\frac{8.029}{2344} \\ r^{13-3}=\frac{8.029}{2344} \\ r^{10}=\frac{8.029}{2344} \\ r=0.567 \end{gathered}$

The common ratio is 0.567.

The expression for fourth term a3 is,

$\begin{gathered} a_3=a_{4-1} \\ a_3=a_0r^{\mleft\{4-1\mright\}} \\ a_3=a_0r^3 \end{gathered}$

Hence,

$\begin{gathered} 2344=a_0(0.567)^3 \\ a_0=\frac{2344}{_{}(0.567)^3} \\ =12859 \end{gathered}$

So, the first term is a0=12859.

So, put a=12859 and r=0.567 in equation (1) to get the explicit formula for the geometric sequence.

$a_{n-1}=12859(0.567)^{n-1}$