Question

How do you find all sets of three consecutive even integers whose sum is between 25 and 45?

Answer 1

Hint: we first need to find the relation between any two consecutive even integers. Then we assume the middle integer and find the rest of the integers of that set. We find the sum and form the mathematical form of the given condition. We solve that inequality to find the sets of even integers.

Complete step-by-step answer:
We have to find three consecutive even integers whose sum is between 25 and 45.
We know that the difference between any two consecutive even integers is 2.
We assume the middle integer of those three integers is $n$.
Therefore, the other two integers will be $n-2$ and $n+2$.
So, the integers are $n-2,n,n+2$.
The sum of these three integers is $\left( n-2 \right)+n+\left( n+2 \right)=3n$.
It is given that the sum of those three consecutive even integers is in between 25 and 45.
From the given condition we can form a mathematical inequality which will give us $25<3n<45$.
We will divide the inequality with 3 and get $\dfrac{25}{3}<\dfrac{3n}{3} < \dfrac{45}{3}$.
The simplified range form for $n$ is $\dfrac{25}{3} < n < 15$.
As $n$ is an even integer, the possible values for $n$ will be $n=10,12,14$.
The sets of three consecutive even integers will be $\left\{ 8,10,12 \right\},\left\{ 10,12,14 \right\},\left\{ 12,14,16 \right\}$.

Note: We need to remember that it is not necessary to assume the middle integer first. This particular process eases the solution to find. The phrase of ‘in between’ omits the possibility of equality.