Question

find $a_2$ and $a_3$.
$a_1 = 5$
$a_n = -6a_{n - 1}$
write your answers as integers or fractions in simplest form.
$a_2 = \square$
$a_3 = \square$

Explanation:

Step1: Find \(a_2\) using the recursive formula

The recursive formula is \(a_n = -6a_{n - 1}\), and we know \(a_1 = 5\). For \(n = 2\), we substitute \(n = 2\) into the formula: \(a_2 = -6a_{2 - 1} = -6a_1\). Since \(a_1 = 5\), we have \(a_2 = -6\times5 = -30\).

Step2: Find \(a_3\) using the recursive formula

For \(n = 3\), we use the recursive formula \(a_n = -6a_{n - 1}\) again. So \(a_3 = -6a_{3 - 1} = -6a_2\). We already found that \(a_2 = -30\), so we substitute that in: \(a_3 = -6\times(-30) = 180\).

Answer:

\(a_2 = -30\)
\(a_3 = 180\)