# Geometric Mean:

#### What is Geometric Mean?

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.

#### The geometric mean of two terms a and b:

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 $$G = \sqrt{ab}$$ $$G = \sqrt{5 \times 10}$$ $$G = \sqrt{50}$$ $$G = 5 \sqrt{2}$$

#### "n" Geometric means between two terms a and b:

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 $$G_1, G_2, G_3$$ $$G_1 = ad = 27 \times \frac{1}{3}$$ $$G_1 = 9$$ $$G_2 = ad^2 = 27 \times \frac{1}{9}$$ $$G_2 = 3$$ $$G_3 = ad^3 = 27 \times \frac{1}{27}$$ $$G_3 = 1$$