Answer:
-2/3
Step-by-step explanation:
The equation of the series:
[tex]a_n = \frac{5-(n-1)}{3} \\a_n = \frac{6-n}{3}\\[/tex]
Where [tex]a_n[/tex] is the nth term and n is the number of terms.
Now let's calculate the partial sum. The equation for the partial sum should be: [tex]S_n = \frac{1}{2}(a_1 +a_n)[/tex]
The first term is 5/3.
The 15th term according to our previous equation is:
[tex]a_{15} = \frac{6-15}{3} \\a_{15} = -3[/tex]
Put all that together we get:
[tex]S_{15} = \frac{1}{2}(\frac{5}{3}+(-3))\\S_{15} = \frac{5}{6}-\frac{3}{2}\\S_{15} = -\frac{2}{3}[/tex]
