The middle term between two other terms of a geometric series is called geometric mean.
Example: In the given geometric series $$ 2, 4, 8, 16...... $$ 4 is the geometric mean of 2 and 8, similarly, 8 is the geometric mean of 4 and 16.
Let "G" is the geometric mean of a and b then it can be written as $$ a, G, b... $$ Hence, $$ \frac{G}{a} = \frac{b}{G} $$ $$ G^2 = ab $$ $$ G = \pm \ \sqrt{ab} $$
Example: Find the geometric mean of two terms 5 and 10 of a GP series?
Solution: Given, a = 5 and b = 10 then geometric mean $$ G = \sqrt{ab} $$ $$ G = \sqrt{5 \times 10} $$ $$ G = \sqrt{50} $$ $$ G = 5 \sqrt{2} $$
Let "n" geometric means between a and b are \(G_1, G_2, G_3....G_n\) then it can be written in geometric series as $$ a, G_1, G_2, G_3....G_n, b $$ Hence total number of terms in this GP series will be (n + 2)
If the common difference of the series is "d" then \((n + 2)^{th}\) term $$ n^{th} \ term = a \ d^{n - 1} $$ then \((n + 2)^{th}\) term $$ b = a \ d^{n + 2 - 1} $$ $$ b = a \ d^{n + 1} $$ $$ \frac{b}{a} = d^{n + 1} $$ $$ d = (\frac{b}{a})^{\frac{1}{n + 1}}...(1) $$ Hence the first term of the series $$ G_1 = ad $$ by putting the value of d from equation (1) $$ G_1 = a \ (\frac{b}{a})^{\frac{1}{n + 1}} $$ second term of the series $$ G_2 = ad^2 $$ $$ G_2 = a \ (\frac{b}{a})^{\frac{2}{n + 1}} $$ similarly \(n^{th}\) term of the series $$ G_n = ad^{n - 1} $$ $$ G_n = a \ (\frac{b}{a})^{\frac{n - 1}{n + 1}} $$
Example: Find out three geometric means between 27 and \(\frac{1}{3}\)?
Solution: Let three geometric means between 27 and \(\frac{1}{3}\) are \(G_1, G_2, G_3\) then we can written as $$ 27, G_1, G_2, G_3, \frac{1}{3} $$ Hence the total number of terms in the series is 5, the first term is 27 and the fifth term is \(\frac{1}{3}\) then $$ T_n = ad^{n - 1} $$ $$ T_5 = ad^{5 - 1} $$ $$ \frac{1}{3} = 27 \times d^4 $$ $$ \frac{1}{81} = d^4 $$ $$ d^4 = (\frac{1}{3})^4 $$ $$ d = \frac{1}{3} $$ Hence the values of \(G_1, G_2, G_3\) $$ G_1 = ad = 27 \times \frac{1}{3} $$ $$ G_1 = 9 $$ $$ G_2 = ad^2 = 27 \times \frac{1}{9} $$ $$ G_2 = 3 $$ $$ G_3 = ad^3 = 27 \times \frac{1}{27} $$ $$ G_3 = 1 $$