1. Find the arithmetic progression $x_i$

Given $x_1=2$ and $x_1,x_2,\dots,x_{100}$ are in A.P. Let the common difference be $d$. Then $x_i=2+(i-1)d.$

The mean of 100 terms of an A.P. is the average of first and last terms: $\frac{x_1+x_{100}}{2}=200.$ So, $\frac{2+x_{100}}{2}=200 \implies x_{100}=398.$

But $x_{100}=2+99d=398,$ therefore $99d=396 \implies d=4.$

Hence $x_i=2+4(i-1)=4i-2.$

2. Compute $y_i$

Given $y_i=i(x_i-i).$ Substitute $x_i=4i-2$: $y_i=i\big((4i-2)-i\big)=i(3i-2)=3i^2-2i.$

So the mean of $y_1,y_2,\dots,y_{100}$ is $\frac{1}{100}\sum_{i=1}^{100}(3i^2-2i).$

3. Use summation formulas

We know $\sum_{i=1}^{100} i = \frac{100\cdot 101}{2}=5050,$ $\sum_{i=1}^{100} i^2 = \frac{100\cdot 101\cdot 201}{6}=338350.$

Thus $\sum_{i=1}^{100}(3i^2-2i)=3\sum_{i=1}^{100}i^2-2\sum_{i=1}^{100}i$ $=3(338350)-2(5050)$ $=1015050-10100=1004950.$

Therefore the mean is $\frac{1004950}{100}=10049.5.$

4. Match with options

$10049.5=10049.50$ So the correct option is B.

5. Comparison with stored answer

Stored correct answer: B

Our derived answer: B

They agree.