# How do you Find the three consecutive even numbers such that the sum of the first and the third is twice the second?

Mar 14, 2018

True for any three consecutive even numbers.

#### Explanation:

Even numbers are of the form $n = 2 k$; $k \in \mathbb{Z}$ ($k$ is an integer).

Let's consider three consecutive even numbers $2 k$, $2 \left(k + 1\right)$, and $2 \left(k + 2\right)$.

"The sum of the first and the third is twice the second".

Algebraically, we would express this as:

$R i g h t a r r o w 2 k + 2 \left(k + 2\right) = 2 \cdot 2 \left(k + 1\right) = 4 \left(k + 1\right)$

Expanding out the parentheses:

$R i g h t a r r o w 2 k + 2 k + 4 = 4 k + 4$

Collecting like terms:

$R i g h t a r r o w 4 k + 4 = 4 k + 4$

$R i g h t a r r o w 0 = 0$

Both sides of the equation reach an equality.

This means that, for any three consecutive even numbers, the sum of the first and the third is always twice the second.