# How do you find the three arithmetic means between 44 and 92?

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#### Explanation

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#### Explanation:

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Alan P. Share
Jan 11, 2017

$56 , 68 , 70$

#### Explanation:

Method 1: Using Arithmetic Sequence Analysis
(I did this version first since the question was asked under "Arithmetic Sequences").

If the initial value, $44$ is denoted as ${a}_{0}$
then we have:
$\textcolor{w h i t e}{\text{XXX}} {a}_{0} = 44$
color(white)("XXX")a_1=?
color(white)("XXX")a_2=?
color(white)("XXX")a_3=?
$\textcolor{w h i t e}{\text{XXX}} {a}_{4} = 92$

but we know that for an arithmetic sequence with initial value ${a}_{0}$ and arithmetic increment $d$,
the ${n}^{t h}$ value in the sequence is:
$\textcolor{w h i t e}{\text{XXX}} {a}_{n} = {a}_{0} + n \cdot d$
$\textcolor{w h i t e}{\text{XXXXXX}}$Some will denote the initial value ${a}_{1}$;
$\textcolor{w h i t e}{\text{XXXXXX}}$in this case the formula becomes:
$\textcolor{w h i t e}{\text{XXXXXXXXX}} {a}_{n} = {a}_{1} + \left(n - 1\right) \cdot d$
$\textcolor{w h i t e}{\text{XXXXXX}}$This seems unnecessarily complicated to me.

so
$\textcolor{w h i t e}{\text{XXX}} {a}_{4} = 92 = 44 + 4 d$
$\textcolor{w h i t e}{\text{XXX}} \rightarrow d = 12$

and therefore
$\textcolor{w h i t e}{\text{XXX}} {a}_{1} = 44 + 12 = 56$
$\textcolor{w h i t e}{\text{XXX}} {a}_{2} = 56 + 12 = 68$
$\textcolor{w h i t e}{\text{XXX}} {a}_{3} = 68 + 12 = 80$

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

Method 2: Just calculate the arithmetic averages
The primary arithmetic mean is the midpoint between $44$ and $92$
$\textcolor{w h i t e}{\text{XXX}} \frac{44 + 92}{2} = 68$

The arithmetic mean between the initial value and the midpoint is
$\textcolor{w h i t e}{\text{XXX}} \frac{44 + 68}{2} = 56$

The arithmetic mean between the midpoint and the final value is
$\textcolor{w h i t e}{\text{XXX}} \frac{68 + 92}{2} = 80$

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