To answer this question we will use the following formulas to compute n arithmetic means between 'a' and 'b':
[tex]\begin{gathered} A_1=a+\frac{b-a}{n+1}, \\ A_2=a+2\cdot\frac{b-a}{n+1}, \\ \ldots \\ A_n=a+n\cdot\frac{b-a}{n+1}\text{.} \end{gathered}[/tex]Substituting n=3, a=-16, and b=4 we get:
[tex]\begin{gathered} A_1=-16+\frac{4-(-16)}{3+1}, \\ A_2=-16+2\cdot\frac{4-(-16)}{3+1}, \\ A_3=-16+3\cdot\frac{4-(-16)}{3+1}\text{.} \end{gathered}[/tex]Simplifying the above results we get:
[tex]\begin{gathered} A_1=-16+\frac{4+16}{4}=-16+\frac{20}{4}=-11, \\ A_2=-16+2\cdot\frac{4+16}{4}=-16+\frac{40}{4}=-6, \\ A_3=-16+3\cdot\frac{4+16}{4}=-16+\frac{60}{4}=-1. \end{gathered}[/tex]Answer: -11, -6, and -1.