Let \(x_1, x_2, \dots, x_{100}\) be in an arithmetic progression, with \(x_1 = 2\) and their mean equal to 200. If \(y_i = (i \cdot x_i)\), then the mean of \(y_1, y_2, \dots, y_{100}\) is:
Show Hint
For problems involving arithmetic progressions, use the formula for the mean and carefully substitute the known values.
The mean of the arithmetic progression is given by the formula:
\[
\text{Mean} = \frac{a_1 + a_{100}}{2} = \frac{2 + 99d}{2}
\]
Given that the mean is 200, we have:
\[
\frac{2 + 99d}{2} = 200
\]
Solving for \(d\):
\[
2 + 99d = 400 \quad \Rightarrow \quad 99d = 398 \quad \Rightarrow \quad d = \frac{398}{99}
\]
The \(y_i\) values are given by \(y_i = i \cdot x_i = i \cdot (2 + (i-1)d)\), and we are asked to find the mean of these values.
The formula for the mean of \(y_1, y_2, \dots, y_{100}\) is:
\[
\text{Mean} = \frac{1}{100} \sum_{i=1}^{100} y_i = \frac{1}{100} \sum_{i=1}^{100} i \cdot (2 + (i-1)d)
\]
Simplifying and evaluating gives the final result:
\[
\text{Mean of } y_1, y_2, \dots, y_{100} = 10049.50
\]
