1, 2 and 3
-1, -2, and -3
6 or -6
answer is : 58,59working: x= first number, x + 1= second numberx + x + 1 =1172x= 116x= 58
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.
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 first integer is 44.
24. The second is 26.
Their sum is 99.
I assume you mean what are the two consecutive integers. Algebraically; X = integers.X + (X + 1) = 2752X + 1 = 2752X = 274X = 137============solution setorSince the consecutive integers differ by 1, then the first number is 274/2 = 137, so the second one is 1 more, 138.
No, it is not possible.
answer is : 58,59working: x= first number, x + 1= second numberx + x + 1 =1172x= 116x= 58
Consecutive means one after another.If the first is a, then the second is a+1.The sum of these isa + a + 1 = 2a+1 = -3772a = -378a = -189So the two numbers are -189 and -188
The sum of the first four non-negative, consecutive, even integers is 20.
10-11-12
-7
Two numbers are consecutive if the second one is the one after the first. Mathematically, if the first number is n then the second is n+1.
You can find that out by trial-and-error, that is, experiment with different numbers. Or, you can call the first number "n", in which case the second number is "n+2", and you solve the equation:n + (n+2) = 32
The first integer is 17.