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Find four consecutive integers whose sum is 114.
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The first integer let's call A The one after that is A+1, then A+2, then A+3... with me so far?
yes :)
Therefore A + A + 1 + A + 2 + A + 3 = 114 can you go from there?
I can try yes
okay tell me what you get.
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well I added 1+2+3
and got 6
okay good. when you add the A's together you should get 4A + 6 = 114. Solve A.
here... \[A + A + 1 + A + 2 + A + 3 = 114\] \[4A = 108\]\[A = 108\div4 = 27\] A2 = A + 1. A3 = A + 2 A4 = A+3 so what is your final answer?
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