Large Arithmetic Progression

The sum of an arithmetic progression can be viewed as the product of the number of terms with the median term. (Indeed, this framing is what gives us the general formula for such a sum.)

In the current problem, the number of terms is given to us: $101$ . The median term can be calculated as $2020 - 50*(0.2) = 2010$ .

Thus the sum of the progression is $2010*101=203010$ .

The sum of the last $101$ terms of the arithmetic progression is given by:

$\begin{aligned} S & = 2020 + (2020 - 0.2) + (2020 - 2(0.2)) + \cdots + (2020 - 100(0.2)) \\ & = 2020 \times 101 - 0.2 (1+2 + 3 + \cdots + 100) \\ & = 2020 \times 101 - \frac 15 \left(\frac {100 \times 101}2\right) \\ & = (2020 - 10)101 = 2010 \times 101 = \boxed{203010} \end{aligned}$

Count backwards!
An AP in reverse is also an AP, with a common difference which is the negative of the common difference of the original AP.
Using $'$ to indicate the reverse AP, we have: $\begin{aligned} a'&=2020, d'=-0.2\\ S'_{101}&=\frac {101}2(2a'+100d)\\ &=\frac {101}2 (2(2020)+100(-0.2))\\ &=101(2020-10)\\ &=203010 \end{aligned}$