-2, -1, 0, 1
if we take the numbers to be: a,a+1,a+2,a+3
Then five times the fourth less twice the second means: 5.(a+3) - 2.(a+1) = 3a+13
So we know that 3a+13 = 7,
so 3a = -6
therefore a = -2.
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.
51
64 65 66 67
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.
They are 14, 16 and 18.
10-11-12
The numbers are 14, 16 and 18.
-7
The let statement is: let the smallest of the three integers be x.
Let x represent the first integer. The second consecutive integer is then x + 1. The equation can be written as x + (x + 1) = 71.
The second integer is 51.The integers are 50, 51, and 52, so therefore the second integer is 51. This problem can be solved using this method:(x) + (x+1) + (x+2) = 1533x + 3 = 1533x = 150x = 50If x = 50, then (x+1) = 51, and (x+2) = 52.
Let the first odd integer be x, then the second one is x + 2. So we have: x + (x + 2) = 52 2x + 2 = 52 subtract 2 to both sides 2x = 50 divide by 2 to both sides x = 25 Thus, the first number is 25. Since consecutive odd integers grow by 2, then the second is 27, the third is 29, the fourth is 31, and the fifth integer is 33. So the sum of the last two integers is 64 ( 31 + 33).