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Question

Statement 1: The variance of first n even natural numbers is n2-14
Statement 2: The sum of first n natural numbers is nn+12 and the sum of squares of first n natural numbers is n(n+1)(2n+1)6


A

Statement 1 is True, Statement 2 is False

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B

Statement 1 is False, Statement 2 is True

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C

Statement 1 is False, Statement 2 is False

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D

Statement 1 is True, Statement 2 is True

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Solution

The correct option is B

Statement 1 is False, Statement 2 is True


Explanation for correct option :

Step-1: Verify statement 1

We know that the sum of the first n natural number is =nn+12

Sum of first n even natural numbers is =2+4+6+...+2n

=21+2+3+...+n=2nn+12=nn+1

The meanx¯ of the first n even natural numbers is =nn+1n

=n+1

We know that the sum of the square first n natural number is =n(n+1)(2n+1)6

Sum of the square of first n even natural numbers =22+42+62+...+2n2

=412+22+...n2=4nn+12n+16=2nn+12n+13

Variance=1ni=1nxi2-x¯2

=1n2nn+12n+13-n+12=2n+12n+13-n+12=n+122n+1-3n+13=n+1n-13=n2-13

So statement 1 is incorrect

Step-2: Verify statement 2

We know that the sum of the first n natural number =nn+12

We know that the sum of the squares first n natural numbers =n(n+1)(2n+1)6

So statement 2 is correct

Hence, option(B) i.e. Statement 1 is False, Statement 2 is True is correct


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