If the mean deviation about the mean of the numbers 1, 2, 3, .........., n, where n is odd, is 5(n + 1)/n, then n is equal to ______________.

Solution

Maths sir · 4 steps

Write n = 2m + 1, so the mean is m + 1 and the deviations pair up neatly on both sides.

1. Find the mean. The mean of 1, 2, …, n is (n+1)/2. Since n is odd, put n = 2m + 1, so the mean is the whole number m + 1.
2. Add the deviations. The numbers below the mean are at distances m, m-1, …, 1, and the numbers above it are at the same distances. So the total is 2(1 + 2 + … + m) = m(m+1).
3. Write it in n. With m = (n-1)/2 and m + 1 = (n+1)/2, the total is (n-1)(n+1)/4. Divide by n and set it equal to the given value.

   (n-1)(n+1)/4n = 5(n+1)/n ⇒ (n-1)/4 = 5

4. So. n - 1 = 20, so n = 21. Check: m = 10 gives a mean deviation of 110/21, and (5 × 22)/21 = 110/21.

The common slip: 41. Counting the deviations on only one side of the mean gives (n²-1)/8n, which leads to 41, not 21.

The idea to keep. For 1 to n with n odd, the mean deviation about the mean is (n²-1)/4n. Deviations on the two sides of the middle term mirror each other, so sum one side and double it.

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