How to get $a_{n+2}-a_{n+1}=\frac{a_n-a_{n+1}}{2}$ from $a_{n+2} = \frac{a_n + a_{n+1}}{2}$

Question

How does one get the below? Can someone break this down for me?

Given:

$$ a_{n+2} = \frac{a_n + a_{n+1}}{2}, $$ for all $n \in \mathbb{N}_{+}$

How do you get:

$$a_{n+2}-a_{n+1}=\frac{a_n-a_{n+1}}{2}$$

Answer 1

hint: $a_{n+2}-a_{n+1} = \dfrac{a_n+a_{n+1}}{2} - a_{n+1}=...$

Answer 2

This relationship can as well be established through a geometric representation, i.e., the $(n+2)$nd term is the middle between the $(n+1)$st and the $n$th.

Thus the distance between two consecutive terms is halved at each new step.

$$\tag{1}|a_{n+2}-a_{n+1}|=\frac12 |a_{n+1} - a_{n}|$$

Besides, there is a sign switch at each step (for example, if $a_{n+1}>a_n$ at a step, then their midpoint $a_{n+2}$ will be less than $a_{n+1}$ at the next step). Thus (1) gives the result:

$$\tag{2} a_{n+2}-a_{n+1}= \ - \ \frac12 (a_{n+1} - a_{n})$$

Remark: this is by no means a shorter proof. I just advocate such a geometrical approach as fruitful as initial (guessing) and final (checking).

Answer 3

This is extremely basic and you definitely need to work on your manipulation!

$$a_{n+2} - a_{n+1} = \frac{a_n+a_{n+1}}{2}-a_{n+1}=\frac{a_n+a_{n+1}-2a_{n+1}}{2}=\frac{a_n-a_{n+1}}{2}$$

I provided the full working here as it appears that you are struggling - but this is usually a single step, since the algebraic manipulation is so trivial.

Please up vote or accept this answer if you are satisfied!