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Mathematics 20 Online
OpenStudy (anonymous):

plse solve this

OpenStudy (anonymous):

This is a series in GP

OpenStudy (anonymous):

with r<1 so Sn=\[\frac{ a }{ 1-r }\] so here a=1/3 and r=1/3

OpenStudy (anonymous):

on substituting, you will yield the result

OpenStudy (anonymous):

as n tends to infinity u can use the formula for Sum to Infinite G.P as Mentioned by Joseph

OpenStudy (anonymous):

okay a= 1/3 r=?

OpenStudy (anonymous):

r = common ratio or difference

OpenStudy (anonymous):

and that is -2/9

OpenStudy (anonymous):

Sorry it is common ratio since it is a GP

OpenStudy (anonymous):

r = t2/t1

OpenStudy (anonymous):

t2 = second term t1 = 1st term

OpenStudy (anonymous):

srry 9/3=3

OpenStudy (anonymous):

1/3

OpenStudy (anonymous):

Yup

OpenStudy (anonymous):

@Joseph91 & @Yahoo! thank u

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