# How do you find three consecutive integers if twice the middle integer is equal to the sum of the first and third?

Oct 1, 2016

It is true for all integers.

#### Explanation:

In any three consecutive integers, twice the middle integer is always equal to the sum of the first and third. Hence this is valid for all integers. Still let us attempt it.

Le the three consecutive integers be $x$, $x + 1$ and $x + 2$. As twice the middle integer is equal to the sum of the first and third, we will have

$2 \times \left(x + 1\right) = x + x + 2$

or $2 x + 2 = 2 x + 2$

Hence, it is true for all integers.

This is true for all consecutive integers.

#### Explanation:

Let's say the lowest integer is equal to $x$. And so the we have 3 numbers: $x , x + 1 , x + 2$. And we know a relation:

$2 \left(x + 1\right) = \left(x\right) + \left(x + 2\right)$

$2 x + 2 = 2 x + 2$

And subtracting $2 x + 2$ from both sides gives us $0 = 0$.

So what does this mean? It means that either there is no solution or no one unique solution.

So let's try it out. Let's do:

$1 , 2 , 3 \to 2 \left(2\right) = 1 + 3 \to 4 = 4$

$7 , 8 , 9 \to 2 \left(8\right) = 7 + 9 \to 16 = 16$

$- 1 , 0 , 1 \to 2 \left(0\right) = - 1 + 1 \to 0 = 0$

This works for all integers.