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The sum of three consecutive even integers is 54. How do you find the integers?

seo-qna
Last updated date: 27th Jul 2024
Total views: 385.5k
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Answer
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Hint: We first try to find the general form of even integers. The form of the integers will be $2k,k\in \mathbb{Z}$. We know that the difference between any two consecutive even integers is always equal to 2. We use that relation to find the integers. We find the additional value of the integers. This gives the value of 54. We solve to find the value of $n$.

Complete step-by-step solution:
We have been given that the sum of three consecutive even integers is 54.
We can define any even integers in the form of $2k,k\in \mathbb{Z}$.
The difference between any two consecutive even integers is always equal to 2.
Now we assume those three consecutive even integers where the integers are $2n-2,2n,2n+2$.
Now if we add these integers the answer would be 54. We express this relation in a mathematical form and get $\left( 2n-2 \right)+2n+\left( 2n+2 \right)=54$.
We now solve the equation to find the solution for the variable $n$.
We separate the variables and the constants of the equation $\left( 2n-2 \right)+2n+\left( 2n+2 \right)=54$.
$\begin{align}
  & \left( 2n-2 \right)+2n+\left( 2n+2 \right)=54 \\
 & \Rightarrow 2n-2+2n+2n+2=54 \\
 & \Rightarrow 6n=54 \\
\end{align}$
Now we divide the both sides of the equation $6n=54$ with 6 to get
$\begin{align}
  & \dfrac{6n}{6}=\dfrac{54}{6} \\
 & \Rightarrow n=9 \\
\end{align}$
The value of the variable $n$ is $n=9$.
We need to find the values of the integers $2n-2,2n,2n+2$.
Multiplying the equation $n=9$ with 2, we get
$\begin{align}
  & n\times 2=9\times 2 \\
 & \Rightarrow 2n=18 \\
\end{align}$
The middle integer of the three numbers is 18.
The smaller number than 18 would be $2n-2=18-2=16$.
The greater number than 18 would be $2n+2=18+2=20$.
Therefore, the integers are $16,18,20$.

Note: We assumed the values where we first assumed the middle number and then found the other two. This helps in eliminating the constant term from the addition. We also could have used the variable $n$ directly to form the equation.