Let's call the smallest of the consecutive odd integers x. Then the other ones are x+2 and x+4 respectively, and your equation becomes 4*(x+x+2)+7 = 7*(x+4) which is equivalent to 8x+15=7x+28 or x=13, so the consecutive integers you're looking for are 13, 15 and 17.
18, 20 and 22
They are 14, 16 and 18.
Their sum is 99.
24. The second is 26.
The 3 consecutive odd positive integers are 7, 9 and 11.
18, 20 and 22
10-11-12
They are 14, 16 and 18.
Three consecutive even integers can be represented by a-2, a & a+2.Five times the first is 5a-10. Three times the second is 3a.5a-10 is 14 more than 3a, so 5a-10 = 3a+14Adding 10 to both sides: 5a = 3a+24Taking 3a from both sides: 2a = 24Therefore a=12The three consecutive even integers are 22, 24 and 26.
The numbers are 14, 16 and 18.
Their sum is 99.
24. The second is 26.
The 3 consecutive odd positive integers are 7, 9 and 11.
n + 2(n+1) + 3(n+2) = 86 n + 2n + 2 + 3n + 6 = 86 6n = 86 - 2 - 6 6n = 78 n = 13 Three consecutive integers are = 13,14,15
the first number is x, the second x+1, third, x+2 and so on.so if the sum of three consecutive integers is 24, the setup would be this:x+x+1+x+2=24if it's consecutive even or odd integers the setup would be x, x+2, x+4,etc.so if the sum of three consecutive odd integers is 21, the setup would be:x+x+2+x+4=21for three or more even consecutive numbers, same setup
This doesn't work for integers. Using 7, 8 and 9 comes up with 146 which is the closest you can get.
If the first of these consecutive integers is x, the second integer would be x + 1, and the third integer would be x + 2.Since the sum of the second and the third integer is 17, we can writex + 1 + x + 2 = 172x + 3 = 172x + 3 - 3 = 17 - 32x = 142x/2 = 14/2x = 7Thus, the consecutive integers are 7, 8, and 9.