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 ?

Let the five consecutive even numbers be 2(x – 2), 2(x – 1), 2x, 2(x + 1) and 2(x + 2)
Their sum = 10x = 440
x = 44 => 2(x – 2) = 84
Second least number of the other set = 2(84) – 121 = 47
This set has its least number as 46.
Sum of the numbers of this set = 46 + 47 + 48 + 49 + 50
= 48 – 2 + 48 – 1 + 48 + 48 + 1 + 48 + 2 => 5(48) = 240