Ask your own question, for FREE!
Mathematics 13 Online
OpenStudy (anonymous):

Find the sum of the following infinite geometric series, if it exists. one-third plus one-ninth plus one-twenty-seventh + one-eighty-first plus and so on

OpenStudy (anonymous):

Find the sum of the following infinite geometric series, if it exists. 4 + 3.2 + 2.56 + 2.048 + …

OpenStudy (zzr0ck3r):

don't use decimals:) 1/3+1/9+1/27....take the second divided by the first term what do you get?

OpenStudy (zzr0ck3r):

wait, is this two questions?

OpenStudy (anonymous):

no

OpenStudy (zzr0ck3r):

how does 4 + 3.2 + 2.56 + 2.048 + relate to 1/3+1/9...?

OpenStudy (anonymous):

sorry those are two different questions

OpenStudy (zzr0ck3r):

1/9)/1/3 = 1/3 so your ratio is 1/3 first term is 1/3 so \[\frac{\frac{1}{3}}{1-\frac{1}{3}}\]

OpenStudy (anonymous):

so 1/3 is the answer

OpenStudy (anonymous):

We observe that the common ratio is the ratio of the successive and the previous term:\[\bf Common \ ratio=\frac{ \frac{ 1 }{ 9 } }{ \frac{ 1 }{ 3 } }=\frac{ \frac{ 1 }{ 27 } }{ \frac{ 1 }{ 9 } }=\frac{ 1 }{ 3 }\]So the geometric series is given by ('a' is the first term; 'r' is the common ratio):\[\bf ar^{n-1}=\frac{ 1 }{ 3 }\left( \frac{ 1 }{ 3 } \right)^{n-1}\]We observe that the series is convergent, i.e. \(\bf |r|<1\), so the sum of the series will be given by:\[\bf S_n=\frac{a}{1-r}\]Can you evaluate the sum? @dedog55

OpenStudy (zzr0ck3r):

\[\frac{\frac{1}{3}}{1-\frac{1}{3}}=\frac{\frac{1}{3}}{\frac{2}{3}}=\frac{1}{3}*\frac{3}{2}=\frac{1}{2}\]

OpenStudy (anonymous):

Find the sum of the following infinite geometric series, if it exists. 4 + 3.2 + 2.56 + 2.048 + …

OpenStudy (zzr0ck3r):

what is the ratio?

OpenStudy (anonymous):

I don't know how to do do this

OpenStudy (zzr0ck3r):

yes you do.

OpenStudy (zzr0ck3r):

divide the second term by the first

OpenStudy (zzr0ck3r):

that is your ratio...

OpenStudy (zzr0ck3r):

what is the ratio?

OpenStudy (anonymous):

1.25

OpenStudy (zzr0ck3r):

3.2/4 = 1.25??????

OpenStudy (anonymous):

ya

OpenStudy (zzr0ck3r):

nope...the top is smaller than the bottom so its less than one

OpenStudy (anonymous):

no its .8

OpenStudy (zzr0ck3r):

yes

OpenStudy (zzr0ck3r):

ok so what is the first term?

OpenStudy (anonymous):

4

OpenStudy (zzr0ck3r):

\[\frac{firstterm}{1-ratio}=answer\]

OpenStudy (anonymous):

is it 3.2

OpenStudy (zzr0ck3r):

\[\frac{4}{1-0.8}=\frac{4}{0.2}=?\]

OpenStudy (zzr0ck3r):

how did you get 3.2?

OpenStudy (anonymous):

I did it wrong

OpenStudy (anonymous):

its 20 hahaha

OpenStudy (anonymous):

What is the r value of the following geometric sequence? 40, 20, 10, 5,…

OpenStudy (zzr0ck3r):

so what part of divide the second term by the first are you not getting?

OpenStudy (zzr0ck3r):

and....we already went over this problem:)

OpenStudy (zzr0ck3r):

sorry not trying to be mean, but you ask the same question over and over with different numbers:) so lets get to why :)

OpenStudy (anonymous):

sorry I sent you the wrong problem

OpenStudy (anonymous):

Find two geometric means between 5 and 135.

OpenStudy (zzr0ck3r):

you should close this and open another.

Can't find your answer? Make a FREE account and ask your own questions, OR help others and earn volunteer hours!

Join our real-time social learning platform and learn together with your friends!
Can't find your answer? Make a FREE account and ask your own questions, OR help others and earn volunteer hours!

Join our real-time social learning platform and learn together with your friends!