9,10,11,12
-7
The sum of the first four non-negative, consecutive, even integers is 20.
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 17.
The integers are -9, -7, -5, -3 (16 is 4 more than 12) The equation is: - (A + A+2) = -4 (A+6) +4 -2A - 2 = -4A - 24 +4 2A - 2 = -20 2A = -18 A = -9
The first integer is 44.
The first two consecutive prime numbers that have a difference of 20 are the numbers 887 and 907.
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
14, 16, 18 and 16, 18, 20
The integers would begin with 10.
The 3 consecutive odd positive integers are 7, 9 and 11.
No, it is not possible.