Question

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

Answer 1

A.P. `-(4)/(3), -1, -(2)/(3),...,4(1)/(3)`

Here, `a = -(4)/(3), d = -1 -((-4)/3) -1 + (4)/(3) = (1)/(3)`

l = `4(1)/(3)`
∴ T_n = l = `4(1)/(3) = a + (n - 1)d`

⇒ `4(1)/(3) = (-4)/(3) + (n - 1) xx (1)/(3)`

∴ `(13)/(3) + (4)/(3) = (1)/(3)(n - 1)`

⇒ `(17)/(3) xx (3)/(1)` = (n – 1)
n – 1 = 17
⇒ n = 17 + 1
= 18
∴ Two middle term are `(18)/(2) and (18)/(2) + 1`
= 9th and 10th term
∴ a₉ + a₁₀ = a + 8d + a + 9d
= 2a + 17d
= `2 xx ((-4)/3) + 17 xx (1)/(3)`

= `(-8)/(3) + (17)/(3)`

= `(9)/(3)`
= 3.