Suppose the middle integer is a. Then the three consecutive integers are a-1, a and a+1. Now, 3*a = (a-1)+(a+1) + 1 So 3a = 2a + 1 => a=1 So the three numbers are 0, 1 and 2
The first integer is 17.
-1, 1 and 3
The integers would begin with 10.
They are 14, 16 and 18.
There are no two such consecutive 'integers'!!Proof: Let the first integer be x.So, the next consecutive integer is x+1.Now, as per the question : x + (x+1) = 24=> 2x=23 (After taking all constants on one side and variable terms on the other).=> x=(23/2) = 11.5So, 11.5 & 12.5 are the required numbers but they are not integers !
The first integer is 17.
The first integer is 44.
-1, 1 and 3
no solution. If you solve for x (where x is the first integer) the answer is a fraction, which is not an integer.
The integers would begin with 10.
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.
They are 14, 16 and 18.
10-11-12
The numbers are 14, 16 and 18.
y, y+2, y+4
There are no two such consecutive 'integers'!!Proof: Let the first integer be x.So, the next consecutive integer is x+1.Now, as per the question : x + (x+1) = 24=> 2x=23 (After taking all constants on one side and variable terms on the other).=> x=(23/2) = 11.5So, 11.5 & 12.5 are the required numbers but they are not integers !
There are no such consecutive integer as is so simple to prove! Suppose the first integer is x. Then the next (consecutive) integer is x+1. Then 2*x + 4*(x+1) = 30 So that 2x + 4x + 4 = 30 6x + 4 = 30 6x = 30 - 4 = 26 x = 26/6 which is NOT an integer.