Let {x} and [x] denote the fractional part of x and
the greatest integer ≤ x respectively of a real
number x. If ∫₀ⁿ { x }dx ,∫₀ⁿ [ x ]dx and 10(n² – n),
( n ∈ N,n > 1 ) are three consecutive terms of a G.P., then n is equal to_____.

Solution

Maths sir · 5 steps

Split [0, n] into unit intervals: each one adds 1/2 to the first integral and k to the second.

1. Fractional part. {x} has period 1, and on [0, 1) it equals x, so each unit interval contributes ∫₀¹ x dx = 1/2. Over n intervals, ∫₀ⁿ {x} dx = n/2.
2. Greatest integer. On [k, k+1), [x] = k, so that interval contributes k. Adding from k = 0 to n - 1 gives ∫₀ⁿ [x] dx = 0 + 1 + … + (n-1) = n(n-1)/2.
3. Use the G.P. For three consecutive terms of a G.P., the square of the middle term equals the product of the outer two.

   (n(n-1)/2)² = n/2 × 10 n(n-1)

4. Cancel and solve. The right side is 5n²(n-1). Since n > 1, n²(n-1) is not zero, so divide both sides by it: (n-1)/4 = 5.
5. So. n - 1 = 20, giving n = 21.

The common slip: 5. Taking ∫₀ⁿ {x} dx as 1/2 (one period only) instead of n/2 leads to n(n-1) = 20, which gives 5, not 21.

The idea to keep. For integrals of [x] or {x} from 0 to an integer, work one unit interval at a time and add. {x} gives 1/2 per interval, and [x] gives k on [k, k+1).

Answer from the NTA final answer key. Solution by PrepWiser.