How many terms of the A.P. $-6, -\dfrac{11}{2}, -5, ….$ make the sum -25?

In above A.P. a = -6 and

$d = -\dfrac{11}{2} - (-6) \\[1em] = -\dfrac{11}{2} + 6 \\[1em] = \dfrac{-11 + 12}{2} \\[1em] = \dfrac{1}{2}.$

Let n terms make the sum = -25.

We know that,

$S_n = \dfrac{n}{2}[2a + (n - 1)d] \\[1em] \Rightarrow -25 = \dfrac{n}{2}[2 \times -6 + (n - 1) \times \dfrac{1}{2}] \\[1em] \Rightarrow -25 = \dfrac{n}{2}[-12 + \dfrac{n}{2} - \dfrac{1}{2}] \\[1em] \Rightarrow -25 \times 2 = n[-12 - \dfrac{1}{2} + \dfrac{n}{2}] \\[1em] \Rightarrow -50 = n[\dfrac{-24 - 1}{2} + \dfrac{n}{2}] \\[1em] \Rightarrow -50 = n[\dfrac{n - 25}{2}] \\[1em] \Rightarrow -50 \times 2 = n^2 - 25n \\[1em] \Rightarrow n^2 - 25n + 100 = 0 \\[1em] \Rightarrow n^2 - 20n - 5n + 100 = 0 \\[1em] \Rightarrow n(n - 20) - 5(n - 20) = 0 \\[1em] \Rightarrow (n - 5)(n - 20) = 0 \\[1em] \Rightarrow n - 5 = 0 \text{ or } n - 20 = 0 \\[1em] \Rightarrow n = 5 \text{ or } n = 20.$

Hence, the number of terms that make the sum -25 are 5 or 20.