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Step-by-step Solution
Step 1: Understand the Problem
We need to find the value of the natural number n (where n > 1 ) such that the three expressions
\int_{0}^{n} \{x\}\,dx,\quad \int_{0}^{n} [x]\,dx,\quad \text{and} \quad 10\bigl(n^2 - n\bigr)
form three consecutive terms of a Geometric Progression (G.P.). Here, \{x\} is the fractional part of x , and [x] is the greatest integer less than or equal to x .
Step 2: Evaluate \displaystyle \int_{0}^{n} \{x\}\,dx
Observe that the fractional part function \{x\} repeats with a period of 1. Over any interval of length 1, from k to k+1 (where k is an integer),
\int_{k}^{k+1} \{x\}\,dx = \int_{0}^{1} x\,dx = \left[ \frac{x^2}{2} \right]_{0}^{1} = \frac{1}{2}.
Since we integrate from 0 to n , and n is a natural number, this integral is the sum of n identical pieces:
\int_{0}^{n} \{x\}\,dx = n \cdot \frac{1}{2} = \frac{n}{2}.
Step 3: Evaluate \displaystyle \int_{0}^{n} [x]\,dx
The greatest integer function [x] (also called the floor function) remains constant on each interval [k, k+1) where k is an integer. So we break the integral into n intervals:
\int_{0}^{n} [x]\ dx = \int_{0}^{1} [x]\ dx + \int_{1}^{2} [x]\ dx + \cdots + \int_{n-1}^{n} [x]\ dx.
For k \in \{0,1,2,\dots,n-1\} , [x]=k on the interval [k, k+1) . Thus,
\int_{k}^{k+1} [x]\ dx = \int_{k}^{k+1} k\ dx = k \times 1 = k.
Summing from k=0 to k=n-1 :
\int_{0}^{n} [x]\ dx = 0 + 1 + 2 + \cdots + (n-1).
The right-hand side is an arithmetic series with sum
1 + 2 + \cdots + (n-1) = \frac{(n-1)n}{2}.
So,
\int_{0}^{n} [x]\ dx = \frac{n(n-1)}{2}.
Step 4: Apply the Geometric Progression (G.P.) Condition
We are told that
\int_{0}^{n} \{x\}\,dx,\quad
\int_{0}^{n} [x]\ dx,\quad
10 \bigl(n^2 - n\bigr)
form three consecutive terms of a G.P. Let these three terms be A , B , and C respectively. Thus,
A = \frac{n}{2}, \quad B = \frac{n(n-1)}{2}, \quad C = 10\bigl(n^2 - n\bigr).
In a geometric progression, B^2 = A \times C . Substituting the values gives
\left(\frac{n(n-1)}{2}\right)^2 = \left(\frac{n}{2}\right)\times 10\bigl(n^2 - n\bigr).
Let us simplify step by step.
Step 4.1: Expand the left-hand side (LHS)
LHS = \left(\frac{n(n-1)}{2}\right)^2 = \frac{n^2(n-1)^2}{4} .
Step 4.2: Expand the right-hand side (RHS)
RHS = \left(\frac{n}{2}\right)\times 10(n^2 - n) = \frac{10n}{2} (n^2 - n) = 5n(n^2 - n) = 5n^3 - 5n^2.
Step 4.3: Equate and simplify
We set LHS = RHS:
\frac{n^2(n-1)^2}{4} = 5n^3 - 5n^2.
Multiply both sides by 4 to clear the denominator:
n^2 (n-1)^2 = 20n^3 - 20n^2.
Factor out n^2 where possible:
n^2 (n-1)^2 = 20n^2(n - 1).
Assuming n \neq 0 , we can divide both sides by n^2 :
(n-1)^2 = 20(n - 1).
If n - 1 \neq 0 , divide both sides by (n - 1) :
n - 1 = 20 \quad \Longrightarrow \quad n = 21.
(Since n>1 by the problem statement, the solution n = 21 is valid.)
Step 5: Conclusion
The only positive integer solution (greater than 1) that satisfies the condition of forming a G.P. is
\boxed{21}.
