48/3 = 16 so the three numbers are 15, 16, 17
48 484 848
The numbers are 46, 48 and 50.
The three consecutive even numbers with a sum of 138 are 44, 46 and 48.
The numbers are 47 and 48. (47 x 48 = 2,256)
48/3 = 16 so the three numbers are 15, 16, 17
48 484 848
x+ (x+1) + (x+2) = 48 3x+3 = 48 3x = 45 x = 15
The numbers are 46, 48 and 50.
The numbers are 48 and 49.
The smallest is 15.
Natural numbers only: Sequence length 1: 48. Sequence length 3: 15, 16, 17. If you allow for negative numbers, the following two are also possible: Sequence length 33: -14, -13, ... , 0, 1, ..., 15, 16, 17. Sequence length 97: -47, -46, ..., 47, 48. There is no other sequence with a total of 48 in the whole number domain. A solution would be: Let m be the first consecutive number. Hence: m + 1 = Second consecutive no. m + 2 = Third consecutive no. So the equation will be: 1st no. + 2nd no. + 3rd no = 48 m + (m + 1) + (m + 2) = 48 3m + 3 = 48 3m = 48 - 3 3m = 45 Dividing both sides by 3 gives: m = 15 by substitution: 15 + 1 = Second consecutive no. = 16 15 + 2 = Third consecutive no. = 17 Hence, the numbers are 15, 16 and 17. :-)
The three consecutive even numbers with a sum of 138 are 44, 46 and 48.
The numbers are 47 and 48. (47 x 48 = 2,256)
The smallest is 14.
The numbers are 15, 16 and 17.
48 1/3, 48 2/3, 48 3/3