1: The variance of first n even natural numbers is n^2 - 1/4 . 2: The sum of first n natural numbers is n(n+1)/2 and the sum of squares of first n nat - Mathematics MCQ AIEEE 2009 2026 | ExamTest.live

mediumMCQAIEEE 20092026Mathematics

1 mark

1: The variance of first n even natural numbers is \frac{n ^{2} - 1}{4} . 2: The sum of first n natural numbers is \frac{n ( n + 1 )}{2} and the sum of squares of first n natural numbers is \frac{n ( n + 1 ) ( 2 n + 1 )}{6} .

A $Statement-1 is true, Statement-2 is true; Statement-2 is a correct explanation for Statement-1$
B $Statement-1 is true, Statement-2 is true; Statement-2 is not a correct explanation for Statement-1$
C $Statement-1 is true, Statement-2 is false$
D $Statement-1 is false, Statement-2 is true$

Solution & Step-by-step Explanation

Numbers: 2, 4, 6, \dots, 2 n .Mean \overset{x}{ˉ} = n + 1 . Variance = \frac{\sum x _{i}^{2}}{n} - (\overset{x}{ˉ})^{2} = \frac{2 ^{2} ( 1 ^{2} + \dots + n ^{2} )}{n} - (n + 1)^{2} = \frac{4}{n} \cdot \frac{n ( n + 1 ) ( 2 n + 1 )}{6} - (n + 1)^{2} = \frac{2 ( n + 1 ) ( 2 n + 1 )}{3} - (n + 1)^{2} = (n + 1) [\frac{4 n + 2 - 3 n - 3}{3}] = \frac{( n + 1 ) ( n - 1 )}{3} = \frac{n ^{2} - 1}{3} .Statement-1 says \frac{n ^{2} - 1}{4}, which is false.Statement-2 is the correct formula for sums, so it is true.

Try it yourself before checking the explanation above.

1: The variance of first n even natural numbers is \frac{n ^{2} - 1}{4} . 2: The sum of first n natural numbers is \frac{n ( n + 1 )}{2} and the sum of squares of first n natural numbers is \frac{n ( n + 1 ) ( 2 n + 1 )}{6} .

Statement-1 is true, Statement-2 is true; Statement-2 is a correct explanation for Statement-1

Statement-1 is true, Statement-2 is true; Statement-2 is not a correct explanation for Statement-1

Statement-1 is true, Statement-2 is false

Statement-1 is false, Statement-2 is true

Practice with timed mock tests and track your performance across Mathematics.