Answer:
Step-by-step explanation:
a = -4/3
d= -1 +4/3 =1/3
last term = 4 1/3 = 13/3
nth term = a +(n-1)d
[tex]\frac{-4}{3}+(n-1)*\frac{1}{3}=\frac{13}{3}\\\\(n-1)*\frac{1}{3}=\frac{13}{3}-\frac{-4}{3}\\\\(n-1)*\frac{1}{3}=\frac{13+4}{3}\\\\(n-1)*\frac{1}{3}=\frac{17}{3}\\\\n-1=\frac{17}{3}*\frac{3}{1}\\\\n-1=17\\\\n=17+1=18[/tex]
Middle terms are 9the term and 10th term
[tex]t_{9}=\frac{-4}{3}+8*\frac{1}{3}=\frac{-4}{3}+\frac{8}{3}=\frac{4}{3}\\\\t_{10}=\frac{-4}{3}+9*\frac{1}{3}=\frac{-4}{3}+\frac{9}{3}=\frac{5}{3}\\\\ t_{9}+t_{10}=\frac{4}{3}+\frac{5}{3}=\frac{9}{3}=3\\\\t_{9}+t_{10}=3[/tex]
