Question

find $a_2$, $a_3$, and $a_4$.
$a_1 = -2$
$a_n = a_{n - 1} + 8$
write your answers as integers or fractions simplest form.
$a_2 = \square$
$a_3 = \square$
$a_4 = \square$

Explanation:

Step1: Find \(a_2\)

To find \(a_2\), we use the recursive formula \(a_n = a_{n - 1}+8\) with \(n = 2\). So we substitute \(a_1=-2\) into the formula:
\(a_2=a_{2 - 1}+8=a_1 + 8\)
Substituting \(a_1=-2\), we get \(a_2=-2 + 8=6\)

Step2: Find \(a_3\)

To find \(a_3\), we use the recursive formula with \(n = 3\). So \(a_3=a_{3 - 1}+8=a_2+8\)
We know \(a_2 = 6\), so substituting we get \(a_3=6 + 8 = 14\)

Step3: Find \(a_4\)

To find \(a_4\), we use the recursive formula with \(n=4\). So \(a_4=a_{4 - 1}+8=a_3+8\)
We know \(a_3 = 14\), so substituting we get \(a_4=14+8=22\)

Answer:

\(a_2 = 6\)
\(a_3 = 14\)
\(a_4 = 22\)