# Nth term in a geometric series

• March 21st 2013, 11:37 PM
bikerboy2442
Nth term in a geometric series
Attachment 27637So for this I have calculated what I believe to be the correct few terms of Q and P Q0=75, Q1=84, Q2=94.08. Always increasing by 12% of what the previous function was. I can't understand why my formula didn't work. I also tried using Qn=Q0*Rn-1 but that provided an answer for n=0that was smaller than 75.

*Edit, I understand that I cant use n-1 because we're starting from n=0 and not n=1 so shouldn't it just be Qn=Q0*Rn?
• March 22nd 2013, 12:14 AM
$Q_0 = 75$
$Q_n= (.12)Q_{n-1} + 75$, in other words, what remains in the system on the previous day plus the new dose of 75mg per day.