Question

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 131 less than double the least number of set x

Answer 1

Let set X be {(2k-4),(2k-2),(2k),(2k+2),(2k+4)}where it is an integer.

According to given condition, 10k=440

Therefore, k=44.

Set X : { (84),(86),(88),(90),(92) }

Now, let the second set be Y.

Set Y : { (m-2),(m-1),(m),(m+1),(m+2) } where m is an integer.

According to the second condition,

=> (m-1)=2*(84) -121

=> m = 48

=> Set Y : { (46),(47),(48),(49),(50) }

Sum of elements in set Y = 240.

Answer 2

The sum of the different set of five consecutive integers is 190.

Given:

The sum of five consecutive even numbers of set x is 440.

To Find:
Find the sum of a different set of five consecutive integers whose second least number is 131 less than double the least number of set x.

Solution:

Let's start by finding the first number in the set x.

Let n be the first even number in the set x. Then, the next four even numbers will be n+2, n+4, n+6, and n+8.

According to the problem, the sum of these five numbers is 440:

n + (n+2) + (n+4) + (n+6) + (n+8) = 440

Simplifying this expression, we get:

5n + 20 = 440

Subtracting 20 from both sides, we get:

5n = 420

Dividing both sides by 5, we get:

n = 84

Therefore, the first even number in set x is 84, and the five consecutive even numbers in the set are 84, 86, 88, 90, and 92. The sum of these numbers is 440, as given in the problem.

Now we need to find a different set of five consecutive integers whose second least number is 131 less than double the least number of set x.

Let m be the least number in this new set of five consecutive integers. Then, the next four consecutive integers will be m+1, m+2, m+3, and m+4.

According to the problem, the second least number (m+1) is 131 less than double the least number (2n):

m+1 = 2n - 131

Substituting n = 84 from the previous calculation, we get:

m+1 = 2(84) - 131 = 37

Subtracting 1 from both sides, we get:

m = 36

Therefore, the five consecutive integers in this new set are 36, 37, 38, 39, and 40.

The sum of these five integers is:

36 + 37 + 38 + 39 + 40 = 190

Therefore, the sum of the different set of five consecutive integers is 190.

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