Find the sum of the two middle most terms of an AP `-(4)/(3), -1, -(2)/(3), …, 4(1)/(3)`.

Explanation

Here, `a=-(4)/(3), d=-1-(-(4)/(3))=(1)/(3)`
Let,` a_(n)=4(1)/(3)`
`rArr (-4)/(3)+(n-1)(1)/(3)=(13)/(3) rArr (n-1)(1)/(3)=(17)/(3)`
`rArr n-1=17 rArr n=18`
Now, the two middle most terms are `a_(9) and a_(10)`.
`:. A_(9)+a_(10)=a+8d+a+9d`
`=2a+17d`
`=2(-(4)/(3))+17((1)/(3))=(-8)/(3)+(17)/(3)=3`