Question

1. the first 5 terms of a sequence are $a_1 = 1$, $a_2 = \frac{5}{3}$, $a_3 = \frac{7}{3}$, $a_4 = 3$, $a_5 = \frac{11}{3}$. based on this information, create an equation that can be used to find the $n^{\text{th}}$ term of the sequence, $a_n$. then, use the equation to find the $16^{\text{th}}$ term of the sequence. equation: $a_n = \square$ $a_{16} = \square$

Explanation:

Step1: Check if it's an arithmetic sequence

First, we check the differences between consecutive terms.
\(a_2 - a_1=\frac{5}{3}-1=\frac{5}{3}-\frac{3}{3}=\frac{2}{3}\)
\(a_3 - a_2=\frac{7}{3}-\frac{5}{3}=\frac{2}{3}\)
\(a_4 - a_3 = 3-\frac{7}{3}=\frac{9}{3}-\frac{7}{3}=\frac{2}{3}\)
\(a_5 - a_4=\frac{11}{3}-3=\frac{11}{3}-\frac{9}{3}=\frac{2}{3}\)
Since the common difference \(d = \frac{2}{3}\) is constant, it's an arithmetic sequence.

Step2: Find the formula for \(a_n\)

The formula for the \(n\)-th term of an arithmetic sequence is \(a_n=a_1+(n - 1)d\).
We know \(a_1 = 1=\frac{3}{3}\) and \(d=\frac{2}{3}\).
Substitute into the formula:
\(a_n=\frac{3}{3}+(n - 1)\times\frac{2}{3}\)
Simplify:
\(a_n=\frac{3 + 2(n - 1)}{3}=\frac{3+2n-2}{3}=\frac{2n + 1}{3}\)

Step3: Find \(a_{16}\)

Substitute \(n = 16\) into \(a_n=\frac{2n + 1}{3}\)
\(a_{16}=\frac{2\times16+1}{3}=\frac{32 + 1}{3}=\frac{33}{3}=11\)

Answer:

Equation: \(a_n=\frac{2n + 1}{3}\)
\(a_{16}=11\)