Inserting Arithmetic Means

To find the total number of inserted arithmetic means $$n$$, we analyze the properties of both types of insertions over the given boundary numbers:

1. Analyze the Geometric Means (GMs) insertion:
Let three geometric means $$G_1, G_2, G_3$$ be inserted between the boundary numbers 3 and 48. This creates a 5-term Geometric Progression (GP) where:
- First term $$b_1 = 3$$
- Fifth term $$b_5 = 48$$

Let the common ratio of this GP be $$r$$. Using the general term formula:
$$b_5 = b_1 \cdot r^4 \implies 48 = 3 \cdot r^4$$
$$r^4 = 16 \implies r = 2$$

The inserted GMs are:
- $$G_1 = 3 \cdot 2 = 6$$
- $$G_2 = 3 \cdot 2^2 = 12$$
- $$G_3 = 3 \cdot 2^3 = 24$$

Thus, the value of the $$2\text{nd}$$ GM ($$G_2$$) is exactly 12.

2. Analyze the Arithmetic Means (AMs) insertion:
Let $$n$$ arithmetic means $$A_1, A_2, \dots, A_n$$ be inserted between 3 and 48. This creates an $$(n+2)$$-term Arithmetic Progression (AP) where:
- First term $$a_1 = 3$$
- Last term $$a_{n+2} = 48$$

Let the common difference of this AP be $$d$$. The formula for the common difference when inserting $$n$$ means is:
$$d = \frac{48 - 3}{n + 1} = \frac{45}{n + 1}$$

The formula for the $$4\text{th}$$ arithmetic mean ($$A_4$$) is:
$$A_4 = a_1 + 4d = 3 + 4\left(\frac{45}{n + 1}\right) = 3 + \frac{180}{n + 1}$$

3. Equate the two values to solve for $$n$$:
The problem states that the $$4\text{th}$$ AM is equal to the $$2\text{nd}$$ GM ($$A_4 = G_2$$):
$$3 + \frac{180}{n + 1} = 12$$
$$\frac{180}{n + 1} = 9$$
$$n + 1 = \frac{180}{9}$$
$$n + 1 = 20 \implies n = 19$$

Thus, the total number of inserted arithmetic terms is 19.