# The sum of five consecutive even numbers of set x is 440. Find the sum of a different set of five consecutive integers whose second least number is 121 less than double the least number of set x?

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.

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