The middle term between the two other terms of an arithmetic series is called the arithmetic mean.
Example: Let an arithmetic series is 2, 4, 6, 8, 10,.......
Here "4" is the arithmetic mean of 2 and 6 similarly, "6" is the arithmetic mean of 4 and 8.
Let "M" is the arithmetic mean of a and b then it can be written as $$ a, M, b,..... $$ Hence, $$ M - a = b - M $$ $$ 2M = a + b $$ $$ \bbox[5px,border:1px solid black] { M = \frac{1}{2} \ (a + b) } $$
Example: Find the arithmetic mean of 10 and 20?
Solution: Let a = 10 and b = 20, then arithmetic mean, $$ M = \frac{1}{2} \ (a + b) $$ $$ M = \frac{1}{2} \ (10 + 20) $$ $$ M = \frac{30}{2} = 15 $$
Let n arithmetic means between a and b are \(M_1, M_2, M_3.....M_n\) then it can be written in arithmetic series as $$ a, M_1, M_2, M_3.....M_n, b $$ Hence total number of terms in this Arithmetic series will be (n + 2).
Let the common difference of the series is "d" then \((n + 2)^{th}\) term, $$ n^{th} \ term = a + (n - 1) \ d $$ $$ Then \ (n + 2)^{th} \ term $$ $$ b = a + (n + 2 - 1) \ d $$ $$ b = a + (n + 1) \ d $$ $$ b - a = (n + 1) \ d $$ $$ d = \frac{b - a}{n + 1} $$ Hence first term of the series, $$ M_1 = a + d $$ $$ M_1 = a + \frac{b - a}{n + 1} $$ $$ M_1 = \frac{an + b}{n + 1}....(1) $$ Now the second term of the series, $$ M_2 = a + 2d $$ $$ M_2 = a + 2 \times \frac{b - a}{n + 1} $$ $$ M_2 = \frac{an + a + 2b - 2a}{n + 1} $$ $$ M_2 = \frac{an - a + 2b}{n + 1} $$ $$ M_2 = \frac{a (n - 1) + 2b}{n + 1}.....(2) $$ Similarly \(n^{th}\) term of the series $$ M_n = a + nd $$ $$ M_n = a + n \times \frac{b - a}{n + 1} $$ $$ M_n = \frac{a + bn}{n + 1} $$