Let's call the three consecutive even numbers a, b, and c, where b is the second number and b = a + 2 and c = b + 2, since the difference between any two consecutive even numbers is 2.
According to the problem, one-third of the second number is equal to one-fourth of the third number. Therefore, we can write the equation:
[tex]\[ \frac{b}{3} = \frac{c}{4} \][/tex]
Now, since c is the third number and it is two more than b, we can substitute c with b + 2 in our equation:
[tex]\[ \frac{b}{3} = \frac{b + 2}{4} \][/tex]
To solve for b, we can cross-multiply to get rid of the fractions:
[tex]\[ 4b = 3(b + 2) \][/tex]
[tex]\[ 4b = 3b + 6 \][/tex]
Subtract 3b from both sides of the equation:
[tex]\[ 4b - 3b = 6 \][/tex]
[tex]\[ b = 6 \][/tex]
Now that we have the second number, we can find the first and third numbers using the relationships b = a + 2 and c = b + 2.
First number, a = b - 2 = 6 - 2 = 4.
Third number, c = b + 2 = 6 + 2 = 8.
The three consecutive even numbers are 4, 6, and 8.
Thus, the answer is (A) 4, 6, 8.
