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
Divide the sum of the three consecutive odd integers by 3: -3 /3 = -1. The smallest of these integers will be two less than -1 and the largest will be two more than -1, so the three consecutive odd integers will be -3, -1, and +1.
The GCF of two consecutive numbers is always 1. The GCF of any set of numbers can't be greater than the smallest of the differences between the numbers.
45 if the smallest is a, then the numbers are a, a+2, a+4, a+6, a+8, a+10 a + (a+2) + (a+4) + (a+6) + (a+8) + (a+10) = 300 6a + 30 = 300 6a = 270 a = 45
I assume you wish to know which four consecutive even integers sum to 340. Let the smallest of the integers be x. Then the others are x+2, x+4 and x+6 and x + (x+2) + (x+4) + (x+6) = 340 iff 4x = 328 iff x = 82. So the integers are 82, 84, 86 and 88.
Integers
The smallest is -1
The smallest of seven consecutive even integers whose sum is 700 is 94.
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.
Divide the sum of the three consecutive odd integers by 3: -3 /3 = -1. The smallest of these integers will be two less than -1 and the largest will be two more than -1, so the three consecutive odd integers will be -3, -1, and +1.