Given AP, – 2, - 4, - 6, …, - 100.

Here,

first term, a = - 2

common difference, d = - 4 – (2) = - 2

last term, l = - 100.

We know that, the nth term 'a_n' of an AP from the end is

a_n = l - (n - 1)d

where l is the last term and d is the common difference.

12th term from the end,

a₁₂ = - 100 - (12 - 1) - 2

= - 100 + 22

= - 78

Hence, the 12th term from the end is - 78