Let $\{\mathrm{x}\}$ and $[\mathrm{x}]$ denote the fractional part of x and the greatest integer $\leq \mathrm{x}$ respectively of a real number $x$. if $\int_{0}^{n}\{x\} d x$, $\int_{0}^{n}[x] d x$ and $10\left(n^{2}-n\right)$, $(n \in N, n>1)$ are three consecutive terms of a G.P. then $n$ is equal to ______

JEE Main · 2020 · 04 Sep 2020 · Shift 2 · Definite Integrals

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} d x

\int_{0}^{n} [x] d x

and

10 (n^{2} - n)

(n \in N, n > 1)

are three consecutive terms of a G.P. then

n

is equal to ______

Answer: 21

Over

[0, n]

the fractional-part integral is

\int_{0}^{n} {x} d x = n \cdot \frac{1}{2} = \frac{n}{2}

, and the greatest-integer integral is

\int_{0}^{n} [x] d x = 0 + 1 + \dots + (n - 1) = \frac{n ( n - 1 )}{2}

. For these, together with

10 (n^{2} - n)

, to be three consecutive terms of a G.P., the middle term squared equals the product of the outer two:

(\frac{n ( n - 1 )}{2})^{2} = \frac{n}{2} \cdot 10 n (n - 1)

. Cancelling

n^{2} (n - 1)

, valid for

n > 1

, gives

n - 1 = 20

, so

n = 21

JEE Main · 2020 · 04 Sep 2020 · Shift 2 · Integer answer