Question

Find $3$ consecutive even numbers such that the sum of the first and last numbers exceed the second number by $10$?

Answer 1

Hint: We assume the middle number of the 3 consecutive even numbers. We know the difference between any two consecutive even numbers is 2. We form the mathematical expression and find the solution of the problem from there.

Complete step by step answer:
The difference between any two consecutive even numbers is 2. It is given that there are 3 consecutive even numbers such that the sum of the first and last numbers exceed the second number by 10.We assume the middle number as $x$.Therefore, the other two numbers will be 2 less and 2 greater than $x$.
The other two numbers will be $x-2$ and $x+2$.
Sum of the first and last number will be $x-2+x+2=2x$.
It is greater than $x$ by 10.
Therefore, the mathematical expression becomes $2x-x=10$.
Simplifying we get $x=10$.
Therefore, the consecutive even numbers are $8,10,12$.

Hence, the 3 consecutive even numbers such that the sum of the first and last numbers exceed the second number by 10 are $8,10,12$.

Note:We could also have taken $2n,n\in \mathbb{Z}$ as the general form of the 3 consecutive even numbers but that makes the equations more complicated. But the answer would have been the same in that case too.