The set is 81, 82 and 83. There is also a set of consecutive even integers whose sum is 246. That set is 80, 82 and 84.
There are no such numbers. The only two numbers that sum to 12 and whose product is -84 are irrational, and so cannot be considered as factors of 84.
The numbers are 21 and 63.
Let the numbers be m & m+1 (consecutive) Hence their squares are m^2 & ( m + 1)^2 => m^2 & m^2 + 2m + 2 Hence Their sum is m^2 + m^2 + 2m + 1 = 85 2m^2 + 2m - 84 = 0 m^2 + m - 42 =0 Factor (m + 7)(m - 6) = 0 Hence the numbers are 6 & 7 .
Let the consecutive i9ntegers be ; n, n+1, n+2, n+3 Hence n+(n+1) +(n+2)+(n+3) = 90 Add the LHS 4n + 6 = 90 4n = 84 n = 84/4 = 21 Hence The four integers are , 21,22,23,24,
The numbers are -30, -28 and -26.
The numbers are 82, 84, 86 and 88.
84 + 86 + 88 = 258
26 28 30
Three consecutive numbers whose sum is 84 are 27, 28 and 29 AND 26, 28 and 30.
84+86=170
87
The integers are -30, -28 and -26.
82, 84, 86, 88
27,28,29The sum of which is 84.
The set is 81, 82 and 83. There is also a set of consecutive even integers whose sum is 246. That set is 80, 82 and 84.
If there are three consecutive numbers which sum to 252, you can find the answer by merely dividing 252 by 3. Then take that number and the two on either side of it. So we get 84, then we take it's neighbors 83 and 85 and we're done. 252.