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Mathematics 14 Online
OpenStudy (howard-wolowitz):

Find a general formula for the nth term of the geometric series with:

OpenStudy (howard-wolowitz):

\[a _{5}=1/5.......r=1/2\]

OpenStudy (howard-wolowitz):

@Michele_Laino

OpenStudy (michele_laino):

please use this formula in order to get a_1: \[{a_5} = {a_1}{r^4}\]

OpenStudy (howard-wolowitz):

ok

OpenStudy (howard-wolowitz):

could you show me how to do these but not the answer after you do the work I will figure out the answer?

OpenStudy (howard-wolowitz):

if no then i'll do it

OpenStudy (michele_laino):

here is your step: \[{a_1} = \frac{{{a_5}}}{{{r^4}}} = \frac{{1/5}}{{{{\left( {1/2} \right)}^4}}} = ...?\]

OpenStudy (howard-wolowitz):

ok that would be 16/5 or 3 1/5

OpenStudy (michele_laino):

that's right!

OpenStudy (howard-wolowitz):

so its a=16

OpenStudy (michele_laino):

now, please keep in mind that the general term a_n is given by the subsequent formula: \[{a_n} = {a_1}{r^{n - 1}}\]

OpenStudy (howard-wolowitz):

\[a _{n}=16 \left( 1/2 \right) ^{n-1}\]

OpenStudy (michele_laino):

no, I think: \[{a_n} = {a_1}{r^{n - 1}} = \frac{{16}}{5}\frac{1}{{{2^{n - 1}}}}\]

OpenStudy (michele_laino):

since a_1 = 16/5

OpenStudy (howard-wolowitz):

so 4

OpenStudy (howard-wolowitz):

a =2

OpenStudy (michele_laino):

pleae your answr is: \[{a_n} = {a_1}{r^{n - 1}} = \frac{{16}}{5}\frac{1}{{{2^{n - 1}}}}\]

OpenStudy (howard-wolowitz):

but thats not a optiion

OpenStudy (michele_laino):

I don't see your options

OpenStudy (howard-wolowitz):

well a=4 ...... a=8 ....... a=16 ...... a=2

OpenStudy (howard-wolowitz):

all have (1/2)

OpenStudy (michele_laino):

sorry what is a?

OpenStudy (michele_laino):

and n what is?

OpenStudy (michele_laino):

please, the general term has to depend on n

OpenStudy (howard-wolowitz):

ok so its what then

OpenStudy (michele_laino):

I think that your answer is: \[{a_n} = \frac{{16}}{5}\frac{1}{{{2^{n - 1}}}}\]

OpenStudy (howard-wolowitz):

so its the 16 one kk gotcha thxs

OpenStudy (michele_laino):

thanks!

OpenStudy (howard-wolowitz):

Find an equivalent fraction for the repeating decimal:

OpenStudy (howard-wolowitz):

\[0.218\]

OpenStudy (howard-wolowitz):

hmm maybe 72/330

OpenStudy (michele_laino):

please I can write this: \[0.218 = \frac{{218}}{{1000}} = \frac{{109}}{{500}}\]

OpenStudy (howard-wolowitz):

ok well u got it because a answer was 218/1000

OpenStudy (michele_laino):

since \[\frac{{72}}{{330}} = \frac{{24}}{{110}} = \frac{{12}}{{55}}\] then your answer can not be a right answer

OpenStudy (howard-wolowitz):

well the choices are 72/330

OpenStudy (howard-wolowitz):

218/1000

OpenStudy (howard-wolowitz):

7/330

OpenStudy (howard-wolowitz):

126/330

OpenStudy (howard-wolowitz):

it has to be B

OpenStudy (michele_laino):

please wait are your repeated decimal 1 and 8?

OpenStudy (howard-wolowitz):

yes 0.218

OpenStudy (michele_laino):

ok! then your ratio is: \[\frac{{218 - 2}}{{990}} = \frac{{216}}{{990}} = \frac{{72}}{{330}}\]

OpenStudy (howard-wolowitz):

u said 218/1000 before

OpenStudy (howard-wolowitz):

but ok

OpenStudy (michele_laino):

since I don't understood your question, sorry!

OpenStudy (howard-wolowitz):

its ok dude i dont either i feel dumb hahhaha

OpenStudy (howard-wolowitz):

converge or diverge

OpenStudy (howard-wolowitz):

new question dude sorry

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