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# Algebra Help Needed!

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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

Jun 4, 2021

#1
+117766
+1

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.

Jun 4, 2021
#2
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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.

Guest Jun 4, 2021
#3
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Also, are you saying I'm doing an AP problem in a normal algebra class? That would make me feel better about not being able to solve it.

Guest Jun 4, 2021
#4
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It says that you are composing an answer, Melody. Are you actually there? This has happened a few times and nothing happened. I do need help. Can I repost the question so more people see it?

Guest Jun 4, 2021
#5
+117766
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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}\\

\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

Jun 4, 2021
#6
+1

Thank you so much. I wasn't using decimals so that might have made the problem a bit more tricky to follow.

Guest Jun 4, 2021
#7
+117766
0

I didn't use the decimals either.  I found out quite quickly that decimal would make my answer less accurate, and an exact answer was needed.

Admittedly I did start using  0.75 and 0.25 but I changed them back to fractions when I actually used them.

Melody  Jun 4, 2021
edited by Melody  Jun 4, 2021