# 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

an = 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. 9th and 10th 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.

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