Question

1. in a sequence of numbers, $a_3 = 36$, $a_4 = 49$, $a_5 = 62$, $a_6 = 75$, $a_7 = 88$. based on this information, create an equation that can be used to find the $n^{th}$ term of the sequence, $a_n$. then, use the equation to find the $15^{th}$ term of the sequence. equation: $a_n = \square$ $a_{15} = \square$

Explanation:

Step1: Check if it's an arithmetic sequence

First, we check the common difference between consecutive terms.
For \(a_3 = 36\) and \(a_4 = 49\), the difference is \(49 - 36=13\).
For \(a_4 = 49\) and \(a_5 = 62\), the difference is \(62 - 49 = 13\).
For \(a_5 = 62\) and \(a_6 = 75\), the difference is \(75 - 62=13\).
For \(a_6 = 75\) and \(a_7 = 88\), the difference is \(88 - 75 = 13\).
Since the common difference \(d = 13\) is constant, this is an arithmetic sequence.

Step2: Find the formula for the nth term of an arithmetic sequence

The general formula for the \(n\)-th term of an arithmetic sequence is \(a_n=a_1+(n - 1)d\). But we know \(a_3\), so we can also express it in terms of \(a_3\).
We know that \(a_n=a_3+(n - 3)d\) (because to get from the 3rd term to the \(n\)-th term, we add the common difference \((n - 3)\) times).
We know \(a_3 = 36\) and \(d=13\), so substituting these values in, we get:
\(a_n=36+(n - 3)\times13\)
Simplify the right - hand side:
First, expand \((n - 3)\times13=13n-39\)
Then \(a_n=36 + 13n-39=13n-3\)

Step3: Find the 15th term

Now that we have the formula \(a_n = 13n-3\), to find \(a_{15}\), we substitute \(n = 15\) into the formula:
\(a_{15}=13\times15-3\)
First, calculate \(13\times15 = 195\)
Then \(a_{15}=195 - 3=192\)

Answer:

Equation: \(a_n=\boldsymbol{13n - 3}\)
\(a_{15}=\boldsymbol{192}\)