## Precalculus: Concepts Through Functions, A Unit Circle Approach to Trigonometry (3rd Edition)

The $n^{th}$ term of the sequence is given by the formula: $a_n =6 \ (-2)^{n-1}$ and $a_5=96$
We are given: $a_{1}=6 ; \ r=-2$ The $n^{th}$ term of the sequence is given by the formula: $a_n=a_1r^{n-1}$ $a_n =6 \ (-2)^{n-1}$ Now, the 5th term can be computed by substituting $5$ for $n$: $a_5=6 \ (-2)^{5-1} \\ =6 \ (-2)^4 \\=96$