Let $$x_1, x_2, \ldots, x_{100}$$ be in an arithmetic progression, with $$x_1 = 2$$ and their mean equal to 200. If $$y_i = ix_i - i$$, $$1 \leq i \leq 100$$, then the mean of $$y_1, y_2, \ldots, y_{100}$$ is

Let the common difference of the arithmetic progression be

$$d.$$

Since the mean of the $$100$$ terms is

$$200,$$

we have

$$\frac{x_1+x_{100}}2=200.$$

Given

$$x_1=2,$$

therefore,

$$x_{100}=398.$$

Now,

$$x_{100}=2+99d,$$

so

$$2+99d=398$$

$$99d=396$$

$$d=4.$$

Hence,

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

Therefore,

$$y_i=i(4i-2)-i$$

$$=4i^2-3i.$$

The required mean is

$$\frac1{100}\sum_{i=1}^{100}(4i^2-3i).$$

Using

$$\sum_{i=1}^{100}i=\frac{100\cdot101}{2}=5050,$$

and

$$\sum_{i=1}^{100}i^2=\frac{100\cdot101\cdot201}{6}=338350,$$

we get

$$\sum_{i=1}^{100}(4i^2-3i)$$

$$=4(338350)-3(5050)$$

$$=1353400-15150$$

$$=1338250.$$

Hence, the required mean is

$$\frac{1338250}{100}$$

$$=\boxed{13382.5}.$$