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

Hint:We know that the arithmetic mean is called the average value. The arithmetic mean is obtained by summing two or more numbers or variables and then dividing by the number of numbers or variables. Suppose let’s say we have two numbers ‘a’ and ‘b’ then the arithmetic mean is given by \[\dfrac{{a + b}}{2}\].

Complete step by step solution:
We need to find three arithmetic means between 44 and 93.
Here we take 44 as an initial value and 93 as a final value.
The arithmetic mean between 44 and 92 is
\[\dfrac{{44 + 92}}{2} = \dfrac{{136}}{2} = 68\].
68 is the midpoint value.
The arithmetic mean between the initial value and the midpoint value is
\[\dfrac{{44 + 68}}{2} = \dfrac{{108}}{2} = 56\]
Since we need three numbers we have found two numbers. For the last number we take the
arithmetic mean between the midpoints and final value vale
\[\dfrac{{68 + 92}}{2} = \dfrac{{160}}{2} = 80\]
Thus the three arithmetic mean between 44 and 92 are 68, 56 and 80.

Note: We can solve this using arithmetic progression. We use the formula \[{a_n} = {a_1} + (n - 1)d\], where ‘\[{a_1}\] is the first term and ‘d’ is the common difference and ‘n’ is the term position.
Using the given problem we have
\[\begin{gathered}
{a_1} = 44 \\
{a_2} = ? \\
{a_3} = ? \\
{a_4} = ? \\
{a_5} = 92 \\
\end{gathered} \]
Thus we have first and last term we need to find the three terms in between them.
Now put \[n = 5\] in \[{a_n} = {a_1} + (n - 1)d\]
\[{a_5} = {a_1} + (5 - 1)d\]
Substituting we have.
\[92 - 44 = 4d\]
Rearranging we have,
\[4d = 92 - 44\]
\[4d = 48\]
Divide by 4 on both sides we have
\[d = \dfrac{{48}}{4}\]
\[d = 12\]
We know that the general form of arithmetic progression is
\[a,a + d,a + 2d,a + 3d,.......a + (n - 1)d\]
Then
\[\begin{gathered}
{a_2} = a + d = 44 + 12 = 56 \\
{a_3} = a + 2d = 44 + 2(12) = 68 \\
{a_4} = a + 3d = 44 + 3(12) = 80 \\
\end{gathered} \]
Thus in both the cases we have the same answer.