[tex]\mathbb{SOLUTION:}[/tex]
Since [tex]a_4 = 34[/tex] and [tex]a_1 = 22[/tex], let add [tex]d[/tex] six times [tex]34[/tex] to get to [tex]22[/tex].
- [tex]\boxed{34 + 6d = 22}[/tex]
Solving for d,
- [tex]\begin{gathered} 6d = -34 + 22 \end{gathered}[/tex]
- [tex]\begin{gathered} d = \frac{-12}{6} \end{gathered}[/tex]
- [tex]\begin{gathered} d = -2 \end{gathered}[/tex]
To find [tex]a_1[/tex], subtract d from [tex]a_4[/tex] three times since [tex]d[/tex] should be subtracted [tex](n - 1)[/tex] times to [tex]a_n[/tex] in order to find [tex]a_1[/tex].
- [tex]\begin{gathered} a_1 = a_4 - (n - 1)d \end{gathered}[/tex]
- [tex]\begin{gathered} a_1 = 34 - (3)(2) \end{gathered}[/tex]
- [tex]\begin{gathered} a_1 = 34 + 6 \end{gathered}[/tex]
- [tex]\begin{gathered} a_1 = 40 \end{gathered}[/tex]
To find the [tex]n^{th}[/tex] term, use the formula of general rule
- [tex]\boxed{a_n = a_1 + (n - 1)d}[/tex]
Solving.
[tex]\begin{gathered} a_n = 40 + (n - 1)(-2) \\ a_n = 40 - 2n + 2 \\ a_n = - 2n + 42 \end{gathered}[/tex]
Thus,
[tex]\begin{gathered}\boxed{\begin{array}{l}a_1 = 40 \\ d = - 2 \\ a_n = - 2n + 42 \end{array}}\end{gathered}[/tex]
