Progress your way - 3

The series of numbers forms an arithmetic progression with common difference $d=2$ .

$102,104,...998$

Solving for the number of terms, we have

$a_n=a_1+(n-1)d$

$998=102+(n-1)(2)$

$n=449$

Finally, the sum is

$s=\dfrac{n}{2}(a_1+a_n)=\dfrac{449}{2}(102+998)=$ $\boxed{246950}$

The even numbers between $101$ and $999$ are:

$102,104,106,\ldots,998$

This list of numbers form an arithmetic progression where

$a=102,\;d=2,\;T_n = 998$

Now, to find the sum of these numbers, we need to find the value of $n$ :

$T_n = a+ (n-1)(d)\\ 998 = 102+(n-1)(2)\\ 998=100+2n\\ 2n=898\\ n=449$

Then, the sum can be calculated with this formula:

$S_n = \dfrac{n}{2}(a+T_n)\\ S_{449} = \dfrac{449}{2}(102+998)\\ =\dfrac{449}{2}(1100)\\ =449(550)\\ =\boxed{246950}$