Best Answer

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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The integers are 41, 42 and 43. Also, using consecutive even integers, you find: 40, 42 and 44.

Q: Find three consecutive integers whose sum is 126?

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