2. Use the formula for finding the [tex]$n$[/tex]th term of an arithmetic sequence to estimate the yearly increase in bullying cases starting in 2013.

Given:
[tex]\[ a_n = a_6 = 20172 \][/tex]
[tex]\[ d = \frac{a_n - a_1}{n - k} \][/tex]
[tex]\[ \begin{array}{c}
a_k = a_1 = 1158 \\
n = 6, \ k = 1
\end{array} \][/tex]

(Note: The yearly increase is also the yearly difference)

[tex]\[
\begin{tabular}{|c|c|c|c|c|c|c|}
\hline
Year & 2013 & 2014 & 2015 & 2016 & 2017 & 2018 \\
\hline
Bullying cases & 1158 & -- & -- & -- & -- & 20172 \\
\hline
\end{tabular}
\][/tex]

[tex]\[ d = 3802.8 \][/tex]



Answer :

To determine the yearly increase in bullying cases from 2013 to 2018, we use the formula for the [tex]\( n \)[/tex]th term of an arithmetic sequence.

Given the arithmetic sequence details:
- The term [tex]\( a_n = a_6 = 20172 \)[/tex] for the year 2018.
- The first term [tex]\( a_1 = 1158 \)[/tex] for the year 2013.
- [tex]\( n = 6 \)[/tex] (since [tex]\( a_6 \)[/tex] corresponds to the 6th term, representing 2018).
- [tex]\( k = 1 \)[/tex] (since [tex]\( a_1 \)[/tex] corresponds to the first term, representing 2013).

With the given data, we want to find the common difference [tex]\( d \)[/tex], which represents the yearly increase in bullying cases.

The formula to find the common difference [tex]\( d \)[/tex] in an arithmetic sequence is:
[tex]\[ d = \frac{a_n - a_1}{n - k} \][/tex]

Substituting the given values into the formula:
[tex]\[ a_n = 20172 \][/tex]
[tex]\[ a_1 = 1158 \][/tex]
[tex]\[ n = 6 \][/tex]
[tex]\[ k = 1 \][/tex]

Hence,
[tex]\[ d = \frac{20172 - 1158}{6 - 1} \][/tex]

Calculating the difference in the numerator:
[tex]\[ 20172 - 1158 = 19014 \][/tex]

Next, solving the denominator:
[tex]\[ 6 - 1 = 5 \][/tex]

Now, dividing the difference by the length of the interval:
[tex]\[ d = \frac{19014}{5} = 3802.8 \][/tex]

Thus, the yearly increase in bullying cases is [tex]\( d = 3802.8 \)[/tex]. This means that each year, from 2013 to 2018, the number of bullying cases increased by 3802.8 cases on average.