Finding an average of an average? ...or midpoint of a midpoint if you will.
I have 2 values: 3.64 and 2.18.
I would like to find the value that's 1/4 of the way to 3.64. (as you can tell I'm not very proficient in math)
So far what I've come up with is: $$\frac{\frac{3.64+2.18}{2}+2.18}{2}=2.545$$​​ or $$\frac{\frac{x+y}{2}+y}{2}=2.545$$​​
2.545 is the desired result. My question is if I'm going about this the right way? I feel I might be overcomplicating this, or that my logic is wrong.
 A: 
$$\frac{\frac{x+y}{2}+y}{2}=2.545$$​​

That's correct, and you can also write it as $\cfrac{x+3y}{4}\,$.
Another way to think at it is to take the second value (which is $y$), then add to it $1/4^{th}$ of the difference between the two (which is $x-y$): $\;y+ \cfrac{1}{4}(x-y)\,$ which of course gives the same result.
A: That's a clever way of coming up with a solution and it does work and is valid.  But there is an easier way that is more expandable.
The difference is $3.65 - 2.18 = 1.47$
You want one $1/4$ of that: $1.47/4 = .3675$
And you want that much more than $2.18$ so $2.18 = 2.5475$
To generalize:  $1/n$ of the way between $y, x; y < x$ is $y + \frac{x - y}n = \frac {y(n-1) + x}n$.
Which is basically the same result that you have.  $\frac{\frac {x+y}2 + y}2 = \frac {x + y + 2y}4 = \frac {x + 3y}4$
A: Use a weighted mean:
$$\frac{3}{4}\times2.15+\frac{1}{4}\times3.64 = 2.545. $$
The important thing is that the weights add up to one, i.e. $3/4+1/4 = 1.$
A: If you have two values
$a$ and $b$,
and you want to go
a fraction $f$
between $a$ and $b$,
where 
$0 \le f \le 1$,
and
$f=0$ means you want to be at $a$
and
$f=1$ means you want to be at $b$,
then choose a value of
$fb+(1-f)a
=f(b-a)+a
$.
This expression does what you want:
if $f=0$, it is $a$;
if $f=1$, it is $b$;
if $0 < f < 1$,
it computes a value between $a$ and $b$.
In your case,
when you say
"1/4 of the way to $a$",
I interpret this
as being closer to $b$
than to $a$,
so that $f$
should be 3/4.
The value is thus
$(3/4)b+(1/4)a$
or
$(3/4)(b-a)+a$
or
$(1/4)(a-b)+b$.
These are equivalent ways
of writing the same expression.
Note that if
$f < 0$ or $f > 1$
then a value outside of
the range $[a, b]$
will be computed.
This is often called
"extrapolation".
If $0 \le f \le 1$,
then a value inside 
the range $[a, b]$
will be computed.
This is often called
"interpolation".
