To solve the problem, we need to evaluate the integrals and set up the geometric progression condition. Let's break it down step by step.
### Step 1: Evaluate the integral \(\int_0^n \{x\} \, dx\)
The fractional part function \(\{x\}\) can be expressed as:
- For \(x\) in the interval \([k, k+1)\), \(\{x\} = x - k\).
Thus, we can break the integral into intervals:
\[
\int_0^n \{x\} \, dx = \sum_{k=0}^{n-1} \int_k^{k+1} (x - k) \, dx
\]
Calculating the integral for each interval:
\[
\int_k^{k+1} (x - k) \, dx = \int_k^{k+1} x \, dx - \int_k^{k+1} k \, dx
\]
\[
= \left[ \frac{x^2}{2} \right]_k^{k+1} - k \left[ x \right]_k^{k+1}
\]
\[
= \left( \frac{(k+1)^2}{2} - \frac{k^2}{2} \right) - k \left( (k+1) - k \right)
\]
\[
= \left( \frac{(k^2 + 2k + 1) - k^2}{2} \right) - k
\]
\[
= \left( \frac{2k + 1}{2} \right) - k = \frac{1}{2}
\]
Thus, for \(n\) intervals:
\[
\int_0^n \{x\} \, dx = n \cdot \frac{1}{2} = \frac{n}{2}
\]
### Step 2: Evaluate the integral \(\int_0^n [x] \, dx\)
The greatest integral function \([x]\) can be expressed as:
\[
\int_0^n [x] \, dx = \sum_{k=0}^{n-1} \int_k^{k+1} k \, dx
\]
Calculating the integral for each interval:
\[
\int_k^{k+1} k \, dx = k \left[ x \right]_k^{k+1} = k \cdot (k+1 - k) = k
\]
Thus, for \(n\) intervals:
\[
\int_0^n [x] \, dx = \sum_{k=0}^{n-1} k = \frac{(n-1)n}{2}
\]
### Step 3: Set up the geometric progression condition
We have established that:
- First term \(A = \frac{n}{2}\)
- Second term \(B = \frac{(n-1)n}{2}\)
- Third term \(C = 10(n^2 - n)\)
For these to be in geometric progression, we need:
\[
B^2 = A \cdot C
\]
Substituting the values:
\[
\left( \frac{(n-1)n}{2} \right)^2 = \frac{n}{2} \cdot 10(n^2 - n)
\]
Simplifying both sides:
\[
\frac{(n-1)^2 n^2}{4} = 5n(n^2 - n)
\]
Multiplying through by 4 to eliminate the fraction:
\[
(n-1)^2 n^2 = 20n(n^2 - n)
\]
Expanding both sides:
\[
(n^2 - 2n + 1)n^2 = 20n^3 - 20n^2
\]
\[
n^4 - 2n^3 + n^2 = 20n^3 - 20n^2
\]
Rearranging gives:
\[
n^4 - 22n^3 + 21n^2 = 0
\]
Factoring out \(n^2\):
\[
n^2(n^2 - 22n + 21) = 0
\]
This gives \(n^2 = 0\) or \(n^2 - 22n + 21 = 0\).
### Step 4: Solve the quadratic equation
Using the quadratic formula:
\[
n = \frac{22 \pm \sqrt{22^2 - 4 \cdot 1 \cdot 21}}{2 \cdot 1} = \frac{22 \pm \sqrt{484 - 84}}{2} = \frac{22 \pm \sqrt{400}}{2} = \frac{22 \pm 20}{2}
\]
This gives:
\[
n = \frac{42}{2} = 21 \quad \text{or} \quad n = \frac{2}{2} = 1
\]
### Step 5: Final answer
Since \(n\) must be a natural number greater than 1, we have:
\[
\boxed{21}
\]
