Let {x} and [x] denote the fractional part of | vedclass

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(C) First,we evaluate the integrals:$\int_{0}^{n}\{x\} dx = n \int_{0}^{1} x dx = n \left[ \frac{x^{2}}{2} \right]_{0}^{1} = \frac{n}{2}$.Next,$\int_{

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$21$

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(C) First,we evaluate the integrals:$\int_{0}^{n}\{x\} dx = n \int_{0}^{1} x dx = n \left[ \frac{x^{2}}{2} \right]_{0}^{1} = \frac{n}{2}$.Next,$\int_{0}^{n}[x] dx = \int_{0}^{n} (x - \{x\}) dx = \int_{0}^{n} x dx - \int_{0}^{n} \{x\} dx = \frac{n^{2}}{2} - \frac{n}{2} = \frac{n(n-1)}{2}$.Given that $\frac{n}{2}$,$\frac{n(n-1)}{2}$,and $10n(n-1)$ are in $G.P.$,the square of the middle term equals the product of the extremes:$\left( \frac{n(n-1)}{2} \right)^{2} = \left( \frac{n}{2} \right) \cdot 10n(n-1)$.$\frac{n^{2}(n-1)^{2}}{4} = 5n^{2}(n-1)$.Since $n > 1$,we can divide both sides by $\frac{n^{2}(n-1)}{4}$:$n-1 = 5 \cdot 4 = 20$.Therefore,$n = 21$.

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

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