∫0nxdx=n∫01xdx=n(2x2)01=2n
and ∫0n[x]dx=∫0n(x−x)dx=(2x2)0n−∫0nxdx=2n2−2n
Now, 2n,2n2−n and 10(n2−n) are in Geometric progression
So,(2n2−n)2=2n⋅10(n2−n)
⇒4n2(n−1)2=5.n2(n−1)
⇒n−1=20⇒n=21 (∵n=1)
JEE Main 2020 — Mathematics Calculus
Let x and [x] denote the fractional part of x and the greatest integer ≤x respectively of a real number x. if ∫0nxdx,∫0n[x]dx and 10(n2−n),(n∈N,n>1) are three consecutive terms of a G.P. then n is equal to__
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