Large Geometric Progression

The sum of the last $10$ terms of the geometric progression is given by:

$\begin{aligned} S & = 2020 + \frac {2020}2 + \frac {2020}{2^2} + \cdots + \frac {2020}{2^9} \\ & = 2020 \left(1 + \frac 12 + \frac 1{2^2} + \cdots + \frac 1{2^9} \right) \\ & = 2020 \times \frac {1-\frac 1{2^{10}}}{1-\frac 12} \\ & = 4040 \left(1 - \frac 1{1024} \right) \\ & \approx 4040 - \frac {4040}{1000} \\ & \approx \boxed{4036} \end{aligned}$

Count backwards!
The common ratio of the reverse GP is the reciprocal of the common ratio of the original GP.
Using $'$ to indicate the reverse GP, we have: $\begin{aligned} a'&=2020, r'=\frac 12\\ S'_{10}&=\frac {a'(1-r'^n)}{1-r'}\\ &=\frac {2020(1-(\tfrac 12)^{10})}{1-\tfrac 12}\\ &=4036\end{aligned}$