If the variance of the first n natural numbers is 10 and the variance of the first m even natural numbers is 16, then m + n is equal to_____.

Solution

Maths sir · 4 steps

Doubling every number multiplies the variance by 4, so one formula handles both sets.

1. Standard result. For 1, 2, …, n the mean is (n+1)/2 and the mean of the squares is (n+1)(2n+1)/6. Subtracting the square of the mean from the mean of the squares gives the formula below.

   σ²_(1,2,…,n) = (n+1)(2n+1)/6 - (n+1)²/4 = (n²-1)/12

2. Find n. (n²-1)/12 = 10 gives n² = 121, so n = 11.
3. Find m. The first m even numbers are 2 × (1, 2, …, m). Multiplying every value by 2 multiplies the variance by 2² = 4, so their variance is 4(m²-1)/12 = (m²-1)/3. Setting this equal to 16 gives m² = 49, so m = 7.
4. So. m + n = 7 + 11 = 18.

The idea to keep. Remember that the variance of the first n natural numbers is (n²-1)/12. Scaling every value by k scales the variance by k², and adding a constant does not change it at all.

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