Thy are 8, 10 and 12.
Their sum is three times the middle number.
You've come to the right place. The integers are: 2, 4, and 6
172
x + x + 1 + x + 2 = 3x + 3 = 3 times the middle number
First, multiply the consecutive numbers. Your total will be the highest factor.
Their sum is three times the middle number.
0,6,18,36,60,90,126,168, etc
EVERY three consecutive numbers add to a multiple of 3: Proof: numbers are n, n + 1 and n + 2. The total is 3n + 3 or 3(n + 1) This means that for any three consecutive numbers, the total is 3 times the middle number.
The sum of 3 consecutive whole numbers is always equal to 3 times the middle number in that sequence.
x+(n+1)+3(n+2)= something then solve,
You've come to the right place. The integers are: 2, 4, and 6
6
172
5872345098234783904672083946728390752430689723409687298290843 theres your answer
To count a triplet in a sequence of numbers, look for three consecutive numbers that are the same. Count how many times this pattern occurs in the sequence.
x + x + 1 + x + 2 = 3x + 3 = 3 times the middle number
249