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Find three consecutive even integers such that the sum of the least integer and the middle integer is 30 more than the greatest integer

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Let the first integer be represented by: 2k

the second integer by: 2(k+1)

and the third by: 2(k+2)

Then, 2k + 2(k+1) = 2(k+2) + 30

2k + 2k + 2 = 2k + 4 + 30

4k + 2 = 2k + 34

2k = 32

k = 16

first integer: 2k = 2(16) = 32

second integer: 2(k+1) = 2(16+1) = 2(17) = 34

third integer: 2(k+2) = 2(16+2) = 2(18) = 36

Answer: 32, 34, 36

User Mike Bryant
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