Let x 1 , x 2 , … , x 100 be in an arithmetic progression with x 1 = 2 and their mean equal to 200. If y i = i ( x i - i ) ,1 i 100 , then the mean of y 1 , y 2 , … , y 100 is [2023]

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Solution

Q.
Let $x_{1}, x_{2}, \dots, x_{100}$ be in an arithmetic progression with $x_{1} = 2$ and their mean equal to 200. If $y_{i} = i (x_{i} - i),$ $1 \leq i \leq 100,$ then the mean of $y_{1}, y_{2}, \dots, y_{100}$ is [2023]

1 10101.50

2 10051.50

3 10049.50

4 10100

Ans.
(3)

$Mean = 200 \Rightarrow \frac{\frac{100}{2} (2 \times 2 + 99 d)}{100} = 200$

$\Rightarrow 4 + 99 d = 400 \Rightarrow d = 4$

$y_{i} = i (x_{i} - i) = i (2 + (i - 1) 4 - i) = 3 i^{2} - 2 i$

$Mean = \frac{\sum y_{i}}{100} = \frac{1}{100} \sum_{i = 1}^{100} (3 i^{2} - 2 i)$

$= \frac{1}{100} {\frac{3 \times 100 \times 101 \times 201}{6} - \frac{2 \times 100 \times 101}{2}}$

$= 101 {\frac{201}{2} - 1} = 101 \times 99.5 = 10049.50$

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