1. Key Concepts and Formulas

   • Fractional Part Function $\{x\}$: Defined as $x - [x]$. It is periodic with period 1. For any integer $k$, $\int_0^{kT} f(x) dx = k \int_0^T f(x) dx$.
   • Greatest Integer Function $[x]$: Denotes the greatest integer less than or equal to $x$. It is a step function.
   • Geometric Progression (GP): Three non-zero terms $a, b, c$ are in GP if $b^2 = ac$.

2. Step-by-Step Solution

Step 1: Calculate the first term of the GP, $\int_0^n \{x\} dx$. The fractional part function $\{x\}$ is periodic with period 1. Therefore, we can use the property of definite integrals of periodic functions: $\int_0^n \{x\} dx = n \int_0^1 \{x\} dx$ In the interval $[0, 1)$, $\{x\} = x$. So, $\int_0^1 \{x\} dx = \int_0^1 x dx = \left[ \frac{x^2}{2} \right]_0^1 = \frac{1^2}{2} - \frac{0^2}{2} = \frac{1}{2}$ Thus, the first term is: $\int_0^n \{x\} dx = n \times \frac{1}{2} = \frac{n}{2}$

Step 2: Calculate the second term of the GP, $\int_0^n [x] dx$. The greatest integer function $[x]$ is a step function. We need to integrate it over intervals where it is constant. $\int_0^n [x] dx = \int_0^1 [x] dx + \int_1^2 [x] dx + \int_2^3 [x] dx + \dots + \int_{n-1}^n [x] dx$ In the interval $[k, k+1)$, $[x] = k$. So, $\int_0^1 [x] dx = \int_0^1 0 dx = 0$ $\int_1^2 [x] dx = \int_1^2 1 dx = 1$ $\int_2^3 [x] dx = \int_2^3 2 dx = 2$ $\dots$ $\int_{n-1}^n [x] dx = \int_{n-1}^n (n-1) dx = n-1$ Summing these values, we get the sum of the first $n-1$ non-negative integers: $\int_0^n [x] dx = 0 + 1 + 2 + \dots + (n-1)$ This is an arithmetic series with sum $\frac{(n-1)((n-1)+1)}{2} = \frac{n(n-1)}{2}$. So, the second term is $\frac{n(n-1)}{2}$.

Step 3: Identify the third term of the GP. The third term is given as $10(n^2 - n)$, which can be written as $10n(n-1)$.

Step 4: Apply the condition for terms to be in Geometric Progression. Let the three consecutive terms be $a = \frac{n}{2}$, $b = \frac{n(n-1)}{2}$, and $c = 10n(n-1)$. For these terms to be in GP, we must have $b^2 = ac$. $\left( \frac{n(n-1)}{2} \right)^2 = \left( \frac{n}{2} \right) \cdot (10n(n-1))$ $\frac{n^2 (n-1)^2}{4} = \frac{10n^2 (n-1)}{2}$ $\frac{n^2 (n-1)^2}{4} = 5n^2 (n-1)$

Step 5: Solve the equation for $n$. Rearranging the equation, we get: $\frac{n^2 (n-1)^2}{4} - 5n^2 (n-1) = 0$ Factor out the common term $n^2(n-1)$: $n^2 (n-1) \left( \frac{n-1}{4} - 5 \right) = 0$ We are given that $n \in N$ and $n > 1$. This implies $n \neq 0$ and $n-1 \neq 0$. Therefore, we can equate the remaining factor to zero: $\frac{n-1}{4} - 5 = 0$ $\frac{n-1}{4} = 5$ $n-1 = 20$ $n = 21$ The value $n=21$ satisfies the condition $n \in N, n > 1$.

1. Common Mistakes & Tips

   • When integrating $[x]$, always split the integral into intervals where $[x]$ is constant.
   • When solving equations involving variables that could be zero, factoring is safer than dividing to avoid losing solutions. In this case, the condition $n>1$ simplifies the solving process.
   • Remember the formula for the sum of the first $k$ natural numbers: $\frac{k(k+1)}{2}$.

2. Summary The problem requires calculating two definite integrals involving the fractional part and greatest integer functions, then using the property of geometric progression. The first integral $\int_0^n \{x\} dx$ evaluates to $\frac{n}{2}$ by using the periodicity of $\{x\}$. The second integral $\int_0^n [x] dx$ evaluates to $\frac{n(n-1)}{2}$ by summing the values of $[x]$ over unit intervals. Setting up the GP condition $b^2=ac$ leads to an algebraic equation that, upon solving, yields $n=21$.

3. Final Answer The final answer is $\boxed{21}$.