Arithmetic mean of the numbers is defined as:  \(\dfrac{{\text{Sum of the numbers}}}{{\text{Number of terms}}}\).

If the numbers are in AP, then \(AM = \dfrac{{\text{First term + last term}}}{2}\).

we can also prove that:  \(AM = \left[ {\frac{{{t_k} + {t_{n - k + 1}}}}{2}} \right],\) 

where \({T_k} = \;{k^{th}}\) term form the beginning and \({t_{n{\rm{ }}-{\rm{ }}k{\rm{ }} + }}_1\) is the \({k^{th}}\) term from the end.

The arithmetic mean \(AM\) of any two numbers \(a\) and \(b\)  is given by \(\frac{{a + b}}{2}\). Then the numbers a, \(AM\) and \(b\) will be in AP.

Single Arithmetic Mean: A number \(A\) is said to be the single \(A.M.\) between two given numbers \(a\) and \(b\) provided \(a\), \(A, b\) are in \(A.P.\)

For example, since \(2, 4, 6\) are in \(A.P\), therefore, 4 is the single \(A.M.\) between 2 and 6.

\(n-\)Arithmetic Means:  The numbers \({A_1},{A_2},{\rm{ }}...,{A_n}\) are said to be the \(n\) arithmetic means between two given numbers \(a\) and \(b\) provided

\(a,{A_1}{A_2},{\rm{ }}...,{A_n},b\) are in \(A.P.\)

Here, \(a\) is the first term and \(b\) is the \({\left( {n + {\rm{ }}2} \right)^{th}}\) term of the \(A.P.\) If \(d\) is the common difference of this \(A.P.\), then we have

\(b = a + (n + 2 - 1)d\) gives \(d = \frac{{b - a}}{{n + 1}}\)

Thus, we have

\({A_1} = a + \frac{{b - a}}{{n + 1}}\;\;and\;\;{A_n} = a + \frac{{n(b - a)}}{{n + 1}}\)

Note: \({A_1} + {A_2} +  \ldots  \ldots .. + {A_n} = n\) \(\left( {\frac{{a + b}}{2}} \right)\)