Find the three consecutive even integers such that the sum of the first two integers is the same as the sum of a third integer and \[6\]
A.\[4,6,8\]
B.\[6,8,10\]
C.\[8,10,12\]
D.\[10,12,14\]

Hint: Here we can see that the integers are consecutive even integers and so the difference among them is of \[2\]. So, let the first number be a, and on the above-given basis, we form all the three terms and put it into the given condition and solve to get the required answer.

Complete step by step answer:

Let the first term of the series be a
And so the second term will be \[{\text{a + 2}}\]
And third term will be \[{\text{a + 4}}\]
Now, put it into the equation given as the sum of the first two integers is the same as the sum of the third integer and \[6\]
\[
  {\text{a + (a + 2) = a + 4 + 6}} \\
   \Rightarrow {\text{2a + 2 = a + 10}} \\
   \Rightarrow {\text{a = 8}} \\
 \]
Hence ,
\[
  {\text{a = 8}} \\
  {\text{a + 2 = 10}} \\
  {\text{a + 4 = 12}} \\
 \]
Therefore the three consecutive even integers are 8,10,12.
So option (c) is our correct answer.

Note: An even number is an integer that can be divided by two and remain an integer or has no remainder. An integer that is not an even number is an odd number. On the other hand, an odd number, when divided by two, will result in a non-integer. Since even numbers are integers, negative numbers can be even.