## Precalculus (6th Edition) Blitzer

$a_{1}=16;a_2=-8;a_3=4;a_4=-2;a_5=1$
The general formula to find the nth term of a Geometric sequence is given as: $a_{n}=a_1r^{n-1}$ Here, $a_n=16(-\dfrac{1}{2})^{n-1}$ Plug $n=1,2,3,4,5$ $a_{1}=16(-\dfrac{1}{2})^{1-1}=16;a_2=16(-\dfrac{1}{2})^{2-1}=-8;a_3=16(-\dfrac{1}{2})^{3-1}=4;a_4=16(-\dfrac{1}{2})^{4-1}=-2;a_5=16(-\dfrac{1}{2})^{5-1}=1$ So, the first $5$ terms of the sequence are $a_{1}=16;a_2=-8;a_3=4;a_4=-2;a_5=1$