Find the sum of two middle most terms of the AP

First term of the AP = - (4/3)

Common difference = d = - 1 - (-4/3) = - 1 + (4/3) = 1/3

Last term = 13/3

Since

a_n = a + (n - 1) × d

∴ 13/3 = (-4/3) + (n - 1) × (1/3)

⇒ (13/3) + (4/3) = (n - 1) × (1/3)

⇒ 17/3 = (n - 1) × (1/3)

⇒ 17 = n - 1

⇒ n = 17 + 1

⇒ n = 18

∴ Two middle most terms of the AP are 18/2 and (18/2) + 1 terms, i.e. 9^th and 10^th terms.

So, a₉ = a + (9 - 1) × d

∴ a₉ = (-4/3) + [8 × (1/3)]

⇒ a₉ = (-4/3) + (8/3) = 4/3

Also, a₁₀ = a₉ + d

= (4/3) + (1/3)

= 5/3

Now, a₁₀ + a₉ = (4/3) + (5/3)

= 9/3

= 3

∴ Sum of two middle most terms of the AP is 3.