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
Divide the sum of the three consecutive integers by 3: 138/3 = 46. The smallest of these integers will be one less than 46 and the largest will be one more than 46, so the three consecutive integers will be 45, 46, and 47.
Divide the sum of the three consecutive odd integers by 3: 402/3 = 134. The smallest of these integers will be two less than 134 and the largest will be two more than 134, so the three consecutive odd integers will be 132, 134, and 136.
Divide the sum of the three consecutive odd integers by 3: 225/3 = 75. The smallest of these integers will be two less than 75 and the largest will be two more than 75, so the three consecutive odd integers will be 73, 75, and 77.
Divide the sum of the three consecutive odd integers by 3: 51 / 3 = 17. The smallest of these integers will be two less than 17 and the largest will be two more than 17, so the three consecutive odd integers will be 15, 17, and 19.
Let x be the smallest of the consecutive odd integers. Since consecutive odd integers differ by 2, we havex + (x + 2) + (x + 4) + (x + 6) = -2204x + 12 = -220 (subtract 12 to both sides)4x = -232 (divide by 4 to both sides)x = -58Thus, the four consecutive odd integers whose sum is -220 are -58, -56, -54, and -52.
The smallest is -1
The smallest of seven consecutive even integers whose sum is 700 is 94.
three consecutives numbers: a = smallest a+2 a+4 14 less = -14 than twice the smallest = 2a so... a+a+2+a+4=-14+2a 3a+6=-14+2a 3a-2a=-14-6 a=-20 answer: smallest = -20 greatest = -16
The smallest is 121.
If n is the smallest of the four integers, their sum is 4n+6.
It is -2.
Suppose the smallest integer is A. The next two even numbers are A+2 and A+4. Using the information supplied we can form an equation: 2A - 14 = A + A+2 + A+4 Rearranging: 2A - 14 = 3A + 6 -20 = A So the three integers are -20, -18 and -16.
85
For x, which is the largest integer of nconsecutive positive integers of which the smallest is m:x = m + n - 1
It is 42 because 42+43+44 = 129
59,61,63,65 It is 59
Divide the sum of the three consecutive integers by 3: 138/3 = 46. The smallest of these integers will be one less than 46 and the largest will be one more than 46, so the three consecutive integers will be 45, 46, and 47.