To solve the problem of finding the \((n + 1)\)th geometric mean (G.M.) inserted between 4 and 2916, where \(2n + 1\) G.M.s are inserted, we can follow these steps:
### Step 1: Understand the Problem
We need to insert \(2n + 1\) geometric means between the numbers 4 and 2916. This means we will have a total of \(2n + 2\) terms in the sequence: the first term (4), the \(2n + 1\) G.M.s, and the last term (2916).
### Step 2: Identify the Terms
The terms can be represented as:
\[ 4, G_1, G_2, \ldots, G_n, G_{n+1}, G_{n+2}, \ldots, G_{2n}, 2916 \]
Here, \(G_{n+1}\) is the \((n + 1)\)th geometric mean we want to find.
### Step 3: Use the Property of G.M.s
In a geometric progression (G.P.), the middle term can be found using the relationship:
\[ G_{n+1}^2 = 4 \times 2916 \]
This is because if \(a\) and \(c\) are the first and last terms, then the middle term \(b\) in a G.P. can be found using:
\[ b = \sqrt{ac} \]
### Step 4: Calculate \(G_{n+1}\)
Now we calculate \(G_{n+1}\):
\[
G_{n+1} = \sqrt{4 \times 2916}
\]
### Step 5: Perform the Multiplication
First, calculate \(4 \times 2916\):
\[
4 \times 2916 = 11664
\]
### Step 6: Find the Square Root
Now, we find the square root of 11664:
\[
G_{n+1} = \sqrt{11664}
\]
### Step 7: Simplify the Square Root
Calculating the square root:
\[
\sqrt{11664} = 108
\]
### Conclusion
Thus, the \((n + 1)\)th geometric mean is:
\[
\boxed{108}
\]
