How do you find three consecutive even integers whose sum is 48?

Hint:We first explain the general formula to express an even integer. Then we take the middle integer first according to the general formula. Based on the difference between two consecutives even integers we find the other two integers. We take their sum and equate with 48. We solve the equation to get the integers.

Complete step by step solution:
It’s given that the sum of three consecutive even integers is 48.

The general formula to express an even integer is $2k,k\in \mathbb{Z}$.

Now we assume the three consecutive variables for the even integers.

Let’s assume the middle integer of those three even integers is $2n,n\in \mathbb{Z}$.

We know the consecutive even integers are at a difference of 2 units.

Therefore, the lowest integer out of those three integers is 2 units less than $2n$ and the highest
integer out of those three integers is 2 units greater than $2n$.

Therefore, the first integer is $2n-2$ and the third integer is $2n+2$.

The sum of these integers $2n-2,2n,2n+2$ is 48.

We take the sum of the variables and equate with 48 to get $\left( 2n-2 \right)+2n+\left( 2n+2
\right)=48$.

We perform the binary operation and get
$\begin{align}
& \left( 2n-2 \right)+2n+\left( 2n+2 \right)=48 \\
& \Rightarrow 2n-2+2n+2n+2=48 \\
& \Rightarrow 6n=48 \\
\end{align}$

We need to find the values of the integers.

We divide the both sides of the equation $6n=48$ by 3.
$\begin{align}
& 6n=48 \\
& \Rightarrow \dfrac{6n}{3}=\dfrac{48}{3} \\
& \Rightarrow 2n=16 \\

\end{align}$

We get the value of the integer $2n$ which is the middle integer.

The other two integers are $2n-2=16-2=14$ and $2n+2=16+2=18$.

Therefore, the integers are $14,16,18$.

Note: We took the middle integer first just to make the equation simpler. We can also take the integers as $2n,2n+2,2n+4$. In that case the integers are the same result of $14,16,18$.it’s better to assume the even integers as its general formula.