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If $A$ and $B$ are variances of the $1^{\text {st }}$ ' $n$ ' even nwmbers and $1^{\text {st }}$ ' $n$ ' odd numbers respectively then
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The correct answer is:
$A=B$
We know that variance of n number which are in A.P. whose first terms is 'a' and common difference $d$ is
$$
\begin{aligned}
& \sigma^2=\frac{a^2\left(n^2-1\right)}{12} \\
& \therefore \operatorname{Var}\left(\sigma_1\right)=\frac{2^2\left(n^2-1\right)}{12}=\frac{\left(n^2-1\right)}{3}=A \\
& \operatorname{Var}\left(\sigma_2\right)=\frac{2^2\left(n^2-1\right)}{12}-\frac{n^2-1}{3}=B \\
& \therefore \quad A=B
\end{aligned}
$$
$$
\begin{aligned}
& \sigma^2=\frac{a^2\left(n^2-1\right)}{12} \\
& \therefore \operatorname{Var}\left(\sigma_1\right)=\frac{2^2\left(n^2-1\right)}{12}=\frac{\left(n^2-1\right)}{3}=A \\
& \operatorname{Var}\left(\sigma_2\right)=\frac{2^2\left(n^2-1\right)}{12}-\frac{n^2-1}{3}=B \\
& \therefore \quad A=B
\end{aligned}
$$
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