Nth term in a geometric series

bikerboy2442

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Jan 23, 2013
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Math help.jpgSo 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?
 
You are correct that there is about twelve percent left after twenty-four hours.

However, your continuous-decay formula appears to have been set up as a compounded-growth formula, which is resulting in your "decay" values getting larger (rather than, as "decay" and common sense dictate, getting smaller) over time. Oops! ;)
 
View attachment 2701So 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?
The Q series always has a fresh dose of 75 mg that hasn't had any chance to decay yet. I would definitely solve the P sequence first, then set Q_n = P_n + 75 mg.
 
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