Even integers (...-4, -2, 0, 2, 4, 6, ...) increase by 2

Let x = the smallest even integer
So, x + 2 = the next integer
and x + 4 = the next integer
and x + 6 = the next integer
and x + 8 = the last (GREATEST) integer

We're told that the 5 integers add to -20
So, we can write: x + (x + 2) + (x + 4) + (x + 6) + (x + 8) = -20
Simplify: 5x + 20 = -20
Subtract 20 from both sides to get: 5x = -40
Solve: x = -8

So, the 5 numbers are -8, -6, -4, -2, 0

The greatest is 0.
So, the quantities are EQUAL

Answer: C

Cheers,
Brent

\((x - 4) + (x - 2) + (x) + (x + 2) + (x + 4) = -20\)
\(5x = -20\)
\(x = -4\)

Col. A: \((x + 4) = 0\)
Col. B: \(0\)

Hence, option C

Why is it not D?

I also arrived at the above set, but then on this one, too: 4, 0, -4, -8, -12, in which case A > B.

But then, the numbers selected by you are not consecutive even integers. They form an AP but not consecutive.