The middle term of the AP 11, 8, 5, … , -79 will be :

Here a = 11 and d = 8 – 11 = -3
Let the given AP contain n terms.
$Tn = -79$
$a + (n – 1)d = -79$
$11 + (n – 1) – 3 = -79$
$11 – 3n + 3 = -79$
$14 + 79 = 3n$
$n = 31$
Thus, the given AP has 31 terms.
∴ its middle term$= \frac{1}{2} (31 + 1) = {16}^{th}$ term
∴ 16^th term = 11 + (16 – 1) – 3 = 11 – 45 = -34