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NTA Abhyas 2020
Exam
NTA Abhyas
Year
2020
Subject
Mathematics
Chapter
Statistics

NTA Abhyas 2020 · Mathematics · Statistics

If the variance of the first n natural numbers is 10 and the...

If the variance of the first nn natural numbers is 1010 and the variance of the first mm even natural numbers is 16,16, then the value of m+nm + n is equal to

Correct answer

18

Step-by-step solution

var(1,2,…n)=10⇒12+22+…+n2n−(1+2+…+nn)2=10var\left(1, 2, \ldots n\right) = 10 ⇒ \dfrac{{1}^{2} + {2}^{2} + \ldots + {n}^{2}}{n} - {\left(\dfrac{1 + 2 + \ldots + n}{n}\right)}^{2} = 10
⇒(n+1)(2n+1)6−(n+12)2=10⇒ \dfrac{\left(n + 1\right)\left(2n + 1\right)}{6} - {\left(\dfrac{n + 1}{2}\right)}^{2} = 10
⇒n2−1=120⇒n=11⇒ {n}^{2} - 1 = 120 ⇒ n = 11
var(2,4,6….,2m)=16⇒var(1,2….,m)=4var\left(2, 4, 6 \ldots . , 2m\right) = 16 ⇒ var\left(1, 2 \ldots . , m\right) = 4
⇒m2−1=48⇒m=7⇒m+n=18⇒ {m}^{2} - 1 = 48 ⇒ m = 7 ⇒ m + n = 18
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