Determine the number of terms \(n\) in the geometric series.
a1 = 320; r = 1/4, Sn = 6825/16
a) 9
b) 7
c) 6
d) 5
+118609 I just did a similar one to this, only with an AP. Have a go by yourself.
Show us what you try, even if you don't get far.
I've already tried. I got to 455/256 = 1/4^n but when I converted it into log and inputted it into my calculator, it said it couldn't do the problem. I also got -181999 = 1/4^n but that also wouldn't work when I tried getting the final answer through converting it into log. I hope this makes sense. I was using log to simplify for n in both instances. That's the only way I know how to do it. I'm not even sure if 455/256 and -181999 are the right end numbers. I'm confused and I need help! I hope I've shown you that I tried. Along the way, I also tried different ways of simplifying and canceling out too. I'm stuck.
+118609 this is a geometric progression problem
a1 = 320; r = 1/4, Sn = 6825/16 find n
For a GP
\(S_n=\frac{a(1-r^n)}{1-r}\\ a=320, \quad r=0.25\quad S_n=6825/16 = 426.5625\\~\\ \frac{6825}{16}=\frac{320(1-0.25^n)}{1-0.25}\\ \frac{6825}{16}=\frac{320(1-0.25^n)}{0.75}\\ \frac{6825}{16}*\frac{3}{4}*\frac{1}{320}=1-0.25^n\\ \frac{1365}{16}*\frac{3}{4}*\frac{1}{64}=1-0.25^n\\ \frac{4095}{4096}=1-0.25^n\\ 0.25^n=1-\frac{4095}{4096}\\ 0.25^n=\frac{1}{4096}\\ \frac{1}{4^n}=\frac{1}{4096}\\ 4^n=4096\\ 4^n=4^6\\ n=6\)
LaTex:
S_n=\frac{a(1-r^n)}{1-r}\\
a=320, \quad r=0.25\quad S_n=6825/16 = 426.5625\\~\\
\frac{6825}{16}=\frac{320(1-0.25^n)}{1-0.25}\\
\frac{6825}{16}=\frac{320(1-0.25^n)}{0.75}\\
\frac{6825}{16}*\frac{3}{4}*\frac{1}{320}=1-0.25^n\\
\frac{1365}{16}*\frac{3}{4}*\frac{1}{64}=1-0.25^n\\
\frac{4095}{4096}=1-0.25^n\\
0.25^n=1-\frac{4095}{4096}\\
0.25^n=\frac{1}{4096}\\
\frac{1}{4^n}=\frac{1}{4096}\\
4^n=4096\\
4^n=4^6\\
n=6
Thank you so much. I wasn't using decimals so that might have made the problem a bit more tricky to follow.
