Let x be the middle integer:
3(x+1) = 7(x-1)
3x + 3 = 7x - 7
3x + 10 = 7x
10 = 4x
So the middle one is 10/4 which is not an integer. So there is really no solution.
5872345098234783904672083946728390752430689723409687298290843 theres your answer
You've come to the right place. The integers are: 2, 4, and 6
x+(n+1)+3(n+2)= something then solve,
249
Suppose the middle integer is 2a. Then the smallest is 2a-2 and the biggest is 2a+2. 4 times the smallest is 8a-8 So largest subtracted from the smallest is (8a-8) - (2a+2) = 6a-10 So, 6a-10 = 2*2a = 4a so that 2a = 10 So the integers are 8, 10 and 12.
The 3 consecutive odd positive integers are 7, 9 and 11.
x + (x + 2) + (x + 4) = 5 (x + 4) so 3x + 6 = 5x + 20 ie -2x = 14 so x = -7 and integers are -7, -5 and -3 Sneaky!
98
This equation works out to 2x = x + 11. This is because the three consecutive odd integers can be represented by x, x + 2, and x + 4. From the equation 2x - x = 11, so x = 11. Checking, 2 times 11 = 22 which is 7 more than 15. The three integers are 11, 13 and 15.
x - the smallest x+2 - the middle x+4 - the largest 3x=5+2(x+4) 3x=5+2x+8 3x-2x=5+8 x=13 -the smaller 13+2=15-the middle 13+4=17- the largest
There are no two consecutive even integers, consecutive odd integers, or consecutive integers that satisfy that relationship.
The sum of three consecutive integers is -72