Question

Assertion :The variance of the series $a,a+d,a+2d,a+3d,....a+2nd$ is $\frac{n(n+1)}{3}{d}^2$. Reason: The sum and the sum of squares of first $n$ natural numbers $\frac{n(n+1)}{2}$ and $\frac{n(n+1)(2n+1)}{6}$ respectively

• A. Both Assertion and Reason are correct and Reason is the correct explanation for Assertion
• B. Both Assertion and Reason are correct but Reason is not the correct explanation for Assertion
• C. Assertion is correct but Reason is incorrect
• D. Assertion is incorrect but Reason is correct

Answer 1

The correct option is A Both Assertion and Reason are correct and Reason is the correct explanation for Assertion

Clearly given series is in A.P and no. of terms $=2n+1$
Now mean of this series is, $\overline{x}=\frac{a+a+2nd}{2}=a+nd$
$\therefore$ Variance, ${\sigma }^2=\frac{\sum (x-\overline{x}{)}^2}{2n+1}=\frac{n^2+(n-1{)}^2+........+2^2+1+1+2^2+...........+(n-1{)}^2+n^2}{2n+1}{d}^2$
$=\frac{2\sum n^2}{2n+1}{d}^2=\frac{2n(n+1)(2n+1)}{6(2n+1)}{d}^2=\frac{n(n+1)}{3}{d}^2$