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

Explanation

Here, first term (a)`=-(4)/(3)`, common difference (d)`= -1+(4)/(3)=(1)/(3)`
and the last term `(l)=4(1)/(3)=(13)/(3)`
` :' n`th term of an AP, `l=a_(n)=a+(n-1)d`
`implies (13)/(3)= - (4)/(3)+(n-1)(1)/(3)`
`implies 13= -4+(n-1)` ,
`implies n-1=17`
`implies " " n=18 " " ` [even]
So, the two middle most terms are `((n)/(2))`th and `((n)/(2)+1)`th. i.e., `((18)/(2))`th and `((18)/(2)+1)`th terms
i.e., 9th and 10th terms.
` :. a_(9)=a+8d= -(4)/(3)+8((1)/(3))=(8-4)/(3)=(4)/(3)`
and `a_(10)=a+9d= -(4)/(3)+9((1)/(3))=(9-4)/(3)=(5)/(3)`
So, sum of the two middle most terms `=a_(9)+a_(10)=(4)/(3)+(5)/(3)=(9)/(3)=3 `