Question

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

Answer 1

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]

⇒ `13/3 = -4/3 + (n - 1)1/3`

⇒ 13 = – 4 + (n – 1)

⇒ n – 1 = 17

⇒ n = 18   ...[Even]

So, the two middle most terms are `(n/12)^("th")` and `(n/2 + 1)^("th")`

i.e., `(18/n)^("th")` and `(18/2 + 1)^("th")`terms

i.e., 9^th and 10^th terms.

∴ a₉ = a + 8d

= `- 4/3 + 8(1/3)`

= `(8 - 4)/3`

= `4/3`

And a₁₀ = `-4/3 + 9(1/3)`

= `(9 - 4)/3`

= `5/3`

So, sum of the two middle most terms

= a₉ + a₁₀

= `4/3 + 5/3`

= `9/3`

= 3