Answer:
10 + 12 + 14
Step-by-step explanation:
10 + 12 + 14 = 36
36/2 = 18
18 is directly between 15 and 21
The the three consecutive even integers such that one half of their sum is between 15 and 21 are 10, 12, 14
let
x = first even integer
(x + 2) = second even integer
(x + 4) = third even integer
Sum = x + (x + 2) + (x + 4)
= x + x + 2 + x + 4
= 3x + 6
one half of sum = 1/2(3x + 6)
15 < 1/2(3x + 6) > 21
first part:
15 < 1/2(3x + 6)
15 × 2 < 3x + 6
30 < 3x + 6
30 - 6 < 3x
24 < 3x
x > 24/3
x > 8
The other part:
1/2(3x + 6) > 21
3x + 6 > 21 × 2
3x + 6 > 42
3x > 42 - 6
3x > 36
x > 12
So,
8 < x < 12
The even integer greater than 8 and less than 12 is 10
x = 10
Therefore,
x = 10
(x + 2) = 10 + 2
= 12
(x + 4) = 10 + 4
= 14
Check:
1/2(3x + 6)
= 1/2(3 × 10 + 6)
= 1/2(30 + 6)
= 1/2(36)
= 18
Therefore, the three consecutive even integers are 10, 12 and 14
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