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Let your middle integer be x. Then your 5 consecutive integers will be:
x-2, x-1, x, x+1, and x+2

Sum of all 5 integers:
(x-2) + (x-1) + (x) + (x+1) + (x+2)
= x -2 + x - 1 + x + x + 1 + x + 2
= 5x

Since the sum equals 200, we have:
5x = 200
x = 40

Your smallest integer in terms of x is x - 2, which is:
40 - 2 = 38.

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Find four consecutive odd integers such that the sum of the first three exceeds the fourth by 18?

We have three consecutive odd integers, To make it easy, I am going to choose the one in the middle to be called X. Therefore, the first odd integer will be called (X-2) and the third odd integer will be called (X+2). The fourth odd integer will be called (X+4).So the sum of the first three, is (X-2) + (X) + (X+2) or X+X+X -2 +2 this equals 3X.Then, 3X -18 equals the fourth odd integer which is (X+4), so 3X - 18 = X + 4Subtract an X from each side for 2X - 18 = 4. Add 18 to each side for 2X = 22. Solve for X, X equals 11. So the first consecutive odd integer is 11-2 or 9, the third is 11+2 or 13 and the fourth is X + 4 or 15. To Check, 9 + 11 + 13 = 33. 15 + 18 = 33


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