JEE MAIN - Mathematics (2023 - 11th April Morning Shift - No. 3)
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 $$y_{1}, y_{2}, \ldots, y_{100}$$ is :
10051.50
10049.50
10100
10101.50
Explanation
We have, mean of $x_1, x_2 \ldots \ldots x_{100}=200$
Where, $x_1, x_2 \ldots x_{100}$ are in AP with first term as 2.
$$ \begin{aligned} \text { Mean } & =200 \\\\ & =\frac{\sum\limits_{i=1}^{100} x_i}{100}=200 \end{aligned} $$
$$ \begin{aligned} \frac{100}{2} \times[2 \times 2+99 d] =20000 \\\\ \Rightarrow 4+99 d =400 \\\\ \Rightarrow 99 d =396 \\\\ d =4 \end{aligned} $$
Also,
$$ \begin{aligned} y_i & =i\left(x_i-i\right) \\\\ & =i[2+(i-1) 4-i] \\\\ & =i[3 i-2] \\\\ & =3 i^2-2 i \end{aligned} $$
$$ \begin{aligned} \text { Required mean } & =\frac{\sum\limits_{i=1}^{100} y_i}{100} \\\\ & =\frac{1}{100}\left[\sum_{i=1}^{100}\left(3 i^2-2 i\right)\right] \\\\ & =\frac{1}{100}\left[\frac{3 \times 100 \times 101 \times 201}{6}-2 \times \frac{100 \times 101}{2}\right]\\\\ & =\frac{20301}{2}-101 \\\\ & =10049.50 \end{aligned} $$
Where, $x_1, x_2 \ldots x_{100}$ are in AP with first term as 2.
$$ \begin{aligned} \text { Mean } & =200 \\\\ & =\frac{\sum\limits_{i=1}^{100} x_i}{100}=200 \end{aligned} $$
$$ \begin{aligned} \frac{100}{2} \times[2 \times 2+99 d] =20000 \\\\ \Rightarrow 4+99 d =400 \\\\ \Rightarrow 99 d =396 \\\\ d =4 \end{aligned} $$
Also,
$$ \begin{aligned} y_i & =i\left(x_i-i\right) \\\\ & =i[2+(i-1) 4-i] \\\\ & =i[3 i-2] \\\\ & =3 i^2-2 i \end{aligned} $$
$$ \begin{aligned} \text { Required mean } & =\frac{\sum\limits_{i=1}^{100} y_i}{100} \\\\ & =\frac{1}{100}\left[\sum_{i=1}^{100}\left(3 i^2-2 i\right)\right] \\\\ & =\frac{1}{100}\left[\frac{3 \times 100 \times 101 \times 201}{6}-2 \times \frac{100 \times 101}{2}\right]\\\\ & =\frac{20301}{2}-101 \\\\ & =10049.50 \end{aligned} $$
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