there are three consecutive even integers such that twice the first is 20 more than the second. find the largest integer
let the three numbers be
x
x+2
x+4
2x - (x+2) = 20
x = 22
largest is x+4 = 26
Let's represent the three consecutive even integers using variables.
Let x be the first even integer.
Then the next two consecutive even integers would be x + 2 and x + 4, respectively.
According to the problem, twice the first integer is 20 more than the second:
2x = (x + 2) + 20
Simplifying the equation:
2x = x + 22
Subtracting x from both sides:
2x - x = 22
x = 22
Therefore, the first even integer is 22, the second is 24 (22 + 2), and the third (largest) integer is 26 (22 + 4).
So, the largest integer in this sequence is 26.
To find the largest integer among three consecutive even integers, let's assign variables to each of the integers.
Let's call the first even integer "x". Since the three integers are consecutive even integers, the second even integer would be "x + 2" (adding 2 to get the next even integer), and the third even integer would be "x + 4" (adding 2 to get the next even integer).
According to the given information, twice the first integer (2x) is 20 more than the second integer, which can be expressed as:
2x = (x + 2) + 20
Now, let's solve this equation to find the value of "x".
2x = x + 22
2x - x = 22
x = 22
So, the first even integer (x) is 22, the second even integer is (22 + 2) = 24, and the third even integer is (22 + 4) = 26.
Thus, the largest integer among the three consecutive even integers is 26.