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The standard deviation of $a, a+d, a+2 d, \ldots, a+2 n d$ is
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Verified Answer
The correct answer is:
$\sqrt{\frac{n(n+1)}{3}} d$
Given series,
$$
a, a+d, a+2 d, \ldots, a+2 n d
$$
Here, $\quad a+2 n d=a+(m-1) d$
$$
\begin{array}{rlrl}
\Rightarrow & & 2 n d=(m-1) d \\
\Rightarrow & 2 n=m-1 \\
\Rightarrow & & 2 n+1=m
\end{array}
$$
$$
\begin{aligned}
& \therefore \text { Standard deviation }=\sqrt{\frac{(2 n+1)^2-1}{12} d^2} \\
&=\sqrt{\frac{(2 n+1-1)(2 n+1+1)}{12} d^2} \\
&=\sqrt{\frac{(2 n+2) 2 n}{12} d^2}=\sqrt{\frac{(n+1) n}{3} d}
\end{aligned}
$$
$$
a, a+d, a+2 d, \ldots, a+2 n d
$$
Here, $\quad a+2 n d=a+(m-1) d$
$$
\begin{array}{rlrl}
\Rightarrow & & 2 n d=(m-1) d \\
\Rightarrow & 2 n=m-1 \\
\Rightarrow & & 2 n+1=m
\end{array}
$$
$$
\begin{aligned}
& \therefore \text { Standard deviation }=\sqrt{\frac{(2 n+1)^2-1}{12} d^2} \\
&=\sqrt{\frac{(2 n+1-1)(2 n+1+1)}{12} d^2} \\
&=\sqrt{\frac{(2 n+2) 2 n}{12} d^2}=\sqrt{\frac{(n+1) n}{3} d}
\end{aligned}
$$
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