Let x _1, x _2 ldots ., x _{100} be in an ari | vedclass

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$	ext { Mean }=200$

$\Rightarrow \frac{\frac{100}{2}(2 	imes 2+99 d)}{100}=200$

$\Rightarrow 4+99 d =400$

$\Rightarrow d=4$

$y_i=i(x i-i

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$10049.50$

Vedclass · Answer

$	ext { Mean }=200$

$\Rightarrow \frac{\frac{100}{2}(2 	imes 2+99 d)}{100}=200$

$\Rightarrow 4+99 d =400$

$\Rightarrow d=4$

$y_i=i(x i-i)$

$=i(2+(i-1) 4-i)=3 i^2-2 i$

$	ext { Mean }=\frac{\sum y_i}{100}$

$=\frac{1}{100} \sum \limits_{i=1}^{100} 3 i^2-2 i$

$=\frac{1}{100}\left\{\frac{3 	imes 100 	imes 101 	imes 201}{6}-\frac{2 	imes 100 	imes 101}{2}ight\}$

$=101\left\{\frac{201}{2}-1ight\}=101 	imes 99.5$

$=10049 \cdot 50$

Let $x _1, x _2 \ldots ., x _{100}$ be in an arithmetic progression, with $x _1=2$ and their mean equal to $200$ . If $y_i=i\left(x_i-i\right), 1 \leq i \leq 100$, then the mean of $y _1, y _2$, $y _{100}$ is

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