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` ,br> `n^(2) - 1 = 120`
`n^(2) = 121`
n = 11
and variance of (2,4,6…..2m) = 16
4 `xx` var (1,2,3,……m) = 16
var (1,2,3,......m) = 4
i.e., `m^(2) - 1 = 48`
`m^(2) = 49 implies m = 7`
`:. lambda + 1 = 11` `mu - 2 = 7`
`lambda + m = 19`