If the variance of first `lambda + 1` natural numbers is 10 and variance of the first `mu - 2` even natural numbers. is 16 then `lambda + mu`

Answer

19

Explanation

Let `lambda +1 = n ` and `mu - 2 = m`
Then ATQ. `(1^(2) + 2^(2) +.....n^(2))/(n) - [((1+2+3....n))/(n)]^(2) = 10`
`n^(2) - 1 = 120`
`n^(2) = 12`
n = 11
and variance of (2,4,6,……..2m) = 16
`4 xx var (1,2,3,…….m) = 16`
i.e., ` m^(2) - 1= 48`
`m^(2) = 49 " " rArr" " m = 7`
`therefore " " lambda + 1 = 11 " " mu - 2 = 7`
`lambda + m = 19`