18, 20 and 22
-7
Their sum is 99.
The let statement is: let the smallest of the three integers be x.
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
Let the first of the three odd consecutive integers be x, so that the second of these integers would be x + 2, and the third one would be x + 4. We have: 3x = 2[(x + 2)+ (x + 4)] + 3 3x = 2(2x + 6) +3 3x = 4x + 12 + 3 (subtract 4x to both sides) -x = 15 (multiply by -1 to both sides) x = -15 (the first one) So the integers are -15, -13, and -11. The average of those integers is (-15 + -13 + -11)/2 = -39/2 = -19.5.
-7
10-11-12
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.
Their sum is 99.
Consecutive integers could be thought of as counting numbers in a row. One of them is "lowest" and the next one will be one more than that, and the last one will be one more than the second one. The numbers 7, 8 and 9 and 46, 47 and 48 are each said to be three consecutive integers. You often come across a question that tells you that 3 consecutive integers add up to a value. Example 3 consecutive integers add to 6. What are the integers? Let the first integer be x the second is then x +1 the third x+2 add them x+x+1+x+2 =3x+3 and this would be equal to 6. we then have the equation 3x+3=6 3x=3 x=1 so the first integer is 1, the second would be 2 and the third 3
This doesn't work for integers. Using 7, 8 and 9 comes up with 146 which is the closest you can get.
1, 2 and 3 -1, -2, and -3 6 or -6
Let the first consecutive integer be x. So that:the second integer is x + 1,the third integer is x + 2, andthe fourth integer is x + 3.We have:(x + 1) + (x + 3) = 1322x + 4 = 1322x = 128x = 64 the first integerThus, the four consecutive integers are 64, 65, 66, and 67.
The let statement is: let the smallest of the three integers be x.
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
121
Since we know that the integers are even and consecutive, we can call them x, x+2, x+4, and x+6, with x being the smallest of the four. twice the sum of the second and third can be written as 2(x+2+x+4)=4x+12 the sum of the first and fourth increased by 14 is x+x+6+14=2x+20 Then we can solve 4x+12=2x+20-->2x=8-->x=4 4