123/3 = 41 so integers are 39, 41 and 43
The integers are are 41, 42 and 43.
There is no set of three consecutive integers for 106.
The consecutive integers are 39, 40, and 41. To solve this algebraically, note that consecutive numbers can be indicated by the variables x, x+1, and x+2. For the sum x + (x+1) + (x+2) = 120, 3x +3 = 120 3x= 117 x= 39, followed by 40 and 41 39+40+41 = 120
123/3 - 2, 123/3 and 123/3 + 2 ie 39, 41 and 43.
123/3 = 41 so integers are 39, 41 and 43
The integers are are 41, 42 and 43.
41+42+43=126.
There is no set of three consecutive integers for 106.
The consecutive integers are 39, 40, and 41. To solve this algebraically, note that consecutive numbers can be indicated by the variables x, x+1, and x+2. For the sum x + (x+1) + (x+2) = 120, 3x +3 = 120 3x= 117 x= 39, followed by 40 and 41 39+40+41 = 120
123/3 - 2, 123/3 and 123/3 + 2 ie 39, 41 and 43.
Start with "-3", then add one at a time to get as many consecutive integers as you want.
Let the intgers be n- 1 n , & n+1 . This is consecutive. Hence Adding (n-1) + n + ( n +1) = 126 Add the LHS 3n = 126 Divide both sides by '3' n = 42 Hence n- 1 = 41 & n +1 = 43 So the consecutive integers are 41,42, & 43.
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.
201
117/3 = 39, so the three consective odd integers whose sum is 117 are 37, 39, and 41.
Heading upwards from -3, the four consecutive integers are -3, -1, 1, 3.-3, -1, 1 and 3.