Answer :
To answer this question we will substitute the given values into the formulas to compute n and d.
Substituting Sₙ=-392, a₁=-2, and aₙ=-54 in the first formula we get:
[tex]-392=\frac{n}{2}(-2+(-54)).[/tex]Simplifying the above result we get:
[tex]\begin{gathered} -392=\frac{n}{2}(-56), \\ -392=-28n. \end{gathered}[/tex]Dividing the above equation by -18 we get:
[tex]\begin{gathered} \frac{-392}{-18}=\frac{-18n}{-18}, \\ 14=n. \end{gathered}[/tex]Now, substituting n=14, aₙ=-54, and a₁=-2 in the second formula we get:
[tex]-54=-2+(14-1)d.[/tex]Simplifying the above result we get:
[tex]-54=-2+13d.[/tex]Then:
[tex]\begin{gathered} -52=13d, \\ -4=d. \end{gathered}[/tex]Now, using the second formula, a₁=-2, and d=-4 we get:
[tex]\begin{gathered} a_2=-2+(2-1)(-4)=-2+(-4)=-6, \\ a_3=-2+(3-1)(-4)=-2+2(-4)=-10. \end{gathered}[/tex]Therefore the first three terms of the arithmetic series are:
[tex]-2+-6+-10+\cdots[/tex]Answer: Second option.