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.
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Those three odd integers are 43, 45, and 47.
99, 100, and 101
42, 44, 46.
-2 + -3 + -4 = -9
The numbers are -134, -132 and -130.
Let's represent the three consecutive integers as x, x+1, and x+2. To find their sum, we add these integers together: x + (x+1) + (x+2) = 88. Simplifying the equation gives us 3x + 3 = 88. Subtracting 3 from both sides and then dividing by 3, we find that x = 28. Therefore, the three consecutive integers are 28, 29, and 30, and their sum is indeed 88.
No. There is no set of three consecutive even integers with a sum of 40.
The integers are 16, 17 and 18.
The integers are 19, 20 and 21.
The integers are -37, -36 and -35. Also, using consecutive even integers: -38, -36 and -34.
Those three odd integers are 43, 45, and 47.
Those three odd integers are 43, 45, and 47.
The even integers are 22, 24 and 26.
99, 100, and 101
31, 32 and 33
-9, -10 and -11.
They are -15, -14 and -13.