Let $$\{x\}$$ and $$[x]$$ denote the fractional part of $$x$$ and the greatest integer $$\leq x$$ respectively of a real number $$x$$. If $$\int_0^n \{x\}dx$$, $$\int_0^n [x]dx$$ and $$10(n^2 - n)$$, $$(n \in N, n > 1)$$ are three consecutive terms of a G.P. then $$n$$ is equal to __________

We begin by recalling the two standard notations: the fractional part of a real number $$x$$ is written as $$\{x\}$$, while the greatest integer less than or equal to $$x$$ is written as $$[x]$$. The problem tells us that the three quantities

$$\int_0^{\,n}\{x\}\,dx,\qquad \int_0^{\,n}[x]\,dx,\qquad 10\,(n^2-n)$$

are consecutive terms of a Geometric Progression (G.P.), where $$n$$ is a natural number greater than 1. In a G.P., if the first term is $$A$$, the second is $$B$$, and the third is $$C$$, then by definition we have the common ratio $$r=\dfrac{B}{A}=\dfrac{C}{B}$$, and this gives the key identity

$$B^2 = A\,C.$$

Our task is therefore to compute the two integrals, label them $$A$$ and $$B$$, set up the equation $$B^2 = A\,C$$, and solve for $$n$$.

Step 1: Evaluate $$\displaystyle\int_0^{\,n}\{x\}\,dx$$.

For any integer $$k$$ with $$0\le k\le n-1$$, on the interval $$[k,k+1)$$ we have $$\{x\}=x-k$$. Hence, for that single unit interval,

$$\int_k^{k+1}\{x\}\,dx \;=\;\int_k^{k+1}(x-k)\,dx.$$

Performing the integration,

$$\int_k^{k+1}(x-k)\,dx \;=\;\left[\frac{(x-k)^2}{2}\right]_{x=k}^{x=k+1} =\frac{(1)^2}{2}-\frac{0^2}{2}=\frac12.$$

Because there are exactly $$n$$ such intervals from $$k=0$$ up to $$k=n-1$$, each contributing $$\dfrac12$$, the total integral is

$$\int_0^{\,n}\{x\}\,dx =\sum_{k=0}^{n-1}\frac12 =\frac{n}{2}.$$

Thus we set

$$A=\frac{n}{2}.$$

Step 2: Evaluate $$\displaystyle\int_0^{\,n}[x]\,dx$$.

On the same interval $$[k,k+1)$$, the function $$[x]$$ is the constant value $$k$$. Therefore

$$\int_k^{k+1}[x]\,dx \;=\;\int_k^{k+1}k\,dx =k\,(x)\Big|_{k}^{k+1}=k(1)=k.$$

Summing over all $$k$$ from $$0$$ to $$n-1$$ gives

$$\int_0^{\,n}[x]\,dx =\sum_{k=0}^{n-1}k =\frac{n(n-1)}{2},$$

using the standard formula for the sum of the first $$n-1$$ natural numbers. Hence we set

$$B=\frac{n(n-1)}{2}.$$

Step 3: Identify the third term of the G.P.

The problem already provides the third term:

$$C=10\,(n^2-n).$$

Step 4: Impose the G.P. condition $$B^2=A\,C$$.

Substituting the expressions for $$A$$, $$B$$, and $$C$$, we write

$$\left(\frac{n(n-1)}{2}\right)^2 \;=\;\left(\frac{n}{2}\right)\,\bigl(10\,(n^2-n)\bigr).$$

Step 5: Simplify the equation.

First expand the left‐hand side:

$$\left(\frac{n(n-1)}{2}\right)^2 =\frac{n^2\,(n-1)^2}{4}.$$

Next simplify the right‐hand side:

$$\left(\frac{n}{2}\right)\,\bigl(10\,(n^2-n)\bigr) =\frac{n}{2}\,\bigl(10n(n-1)\bigr) =5n\,(n^2-n) =5n^2\,(n-1).$$

Equating the two expressions, we have

$$\frac{n^2\,(n-1)^2}{4}=5n^2\,(n-1).$$

Step 6: Cancel the common non-zero factors.

Because $$n>1$$, both $$n$$ and $$n-1$$ are positive, so we may divide both sides by $$n^2(n-1)$$ without fear:

$$\frac{(n-1)}{4}=5.$$ Multiplying through by $$4$$ gives $$(n-1)=20.$$

Step 7: Solve for $$n$$.

$$n = 20+1 = 21.$$

Since $$n=21$$ indeed satisfies $$n\in\mathbb N$$ and $$n>1$$, it is the required value.

So, the answer is $$21$$.