Let n be an odd natural number such that the variance of 1, 2, 3, 4, ......, n is 14. Then n is equal to _____________.

Solution

Maths sir · 4 steps

For the first n natural numbers the variance is always (n²-1)/12, so set that equal to 14.

1. Two averages. The mean of 1, 2, …, n is (n+1)/2. The mean of their squares is 1/nΣ k² = (n+1)(2n+1)/6.
2. Subtract. Variance is the mean of the squares minus the square of the mean. Take (n+1)/12 common and simplify the bracket: 2(2n+1) - 3(n+1) = n - 1.

   σ² = (n+1)(2n+1)/6 - (n+1)²/4 = (n+1)(n-1)/12 = (n²-1)/12

3. Solve. (n²-1)/12 = 14 gives n² - 1 = 168, so n² = 169 and n = 13. The value -13 is rejected because n is a natural number, and 13 is odd as required.
4. So. n = 13.

The idea to keep. Remember the ready result: the variance of 1, 2, …, n is (n²-1)/12. It comes from the mean of squares minus the square of the mean, and it saves you the algebra in the exam.

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