Sequences and Series - Geometric Mean (G.M.)

Insert 3 geometric means between 1 and 256.

Let $G_1, G_2, G_3$ be the three geometric means between $a = 1$ and $b = 256$. The sequence $1, G_1, G_2, G_3, 256$ forms a G.P. Here, the number of inserted means $n = 3$. The common ratio $r$ is: $r = \left( \frac{b}{a} \right)^{\frac{1}{n+1}} = \left( \frac{256}{1} \right)^{\frac{1}{3+1}} = (256)^{\frac{1}{4}}$ Since $256 = 4^4$, we have: $r = (4^4)^{\frac{1}{4}} = 4$ Now find the means: $G_1 = ar = 1 \times 4 = 4$ $G_2 = ar^2 = 1 \times 4^2 = 16$ $G_3 = ar^3 = 1 \times 4^3 = 64$ The three G.M.s are 4, 16, and 64.

To insert $n$ means, we first find the common ratio $r$ using the formula $r = (b/a)^{1/(n+1)}$. Then each mean $G_k$ is found by $a \cdot r^k$.

If the Arithmetic Mean (A.M.) of two positive numbers is 10 and their Geometric Mean (G.M.) is 8, find the numbers.

Let the two numbers be $a$ and $b$. Given: $A.M. = \frac{a+b}{2} = 10 \implies a+b = 20 \quad \text{---(i)}$ $G.M. = \sqrt{ab} = 8 \implies ab = 64 \quad \text{---(ii)}$ We use the identity $(a-b)^2 = (a+b)^2 - 4ab$: $(a-b)^2 = (20)^2 - 4(64)$ $(a-b)^2 = 400 - 256 = 144$ $a-b = \pm 12 \quad \text{---(iii)}$ Case 1: $a+b = 20$ and $a-b = 12$: $\begin{array}{r} a + b = 20 \\ a - b = 12 \\ \hline 2a = 32 \end{array}$ So $a = 16$ and $b = 4$. Case 2: $a+b = 20$ and $a-b = -12$ gives $a = 4$ and $b = 16$.

Using the definitions of A.M. and G.M., we set up a system of equations. Solving for $a+b$ and $ab$ allows us to find $a-b$, leading to the individual values.