Answer: Option D
:
D
The sum of even, odd, even is odd and the sum of odd, even, odd is even. To get an even sum, the three consecutive integers are odd, even, and odd (or) should start with odd. Hence we have 1 + 2+ 3 = 6 which is 6 less than 3 + 4 + 5 = 12 since 3 is 2 more than 1, 4 is 2 more than 2 and 5 is 2 more than 3. Therefore 5 + 6 + 7 is 6 more than 3 + 4 + 5. Hence we get all multiples of 6 less than 50. They are 6,12,18,24,30,36,42,48.
Hence option (d) is the correct choice.
Alternate Solution:
Let the three consecutive integers be (a-1), (a), (a+1).
(a - 1) + a + (a + 1) = 3a = 50. So all even multiples of 3 which are less than 50, can be written in the form of 3 consecutive integers.