We can use the same method to solve this problem by representing the five consecutive even numbers as:

2(x – 2), 2(x – 1), 2x, 2(x + 1), 2(x + 2)

Their sum is given as 440, so we can write:

2(x – 2) + 2(x – 1) + 2x + 2(x + 1) + 2(x + 2) = 440

Simplifying, we get:

10x + 4 = 440

10x = 436

x = 43.6

Since we need to find five consecutive integers with the second least number being 121 less than double the least number of set x, we can write:

2(x – 2) – 121 = m

Substituting the value of “x”, we get:

2(41.6) – 121 = m

m = 47

So the five consecutive integers whose second least number is 121 less than double the least number in set x are 46, 47, 48, 49, and 50.

The sum of these integers is:

46 + 47 + 48 + 49 + 50 = 240

Therefore, the sum of the different set of five consecutive integers is 240.