Question 338482
A bit tricky to explain this one, but recognize that you are adding 10, then half of that, 5, then half of that, 2.5, etc.

Whenever you add half as much, there's usually a (1/2)^n factor involved somewhere...here is a problem like that...

I get a-sub-n = 20 - 20(1/2)^(n-1) or 20(1 - 1/[2^(n-1)])...see if that works for you...