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plse solve this
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This is a series in GP
with r<1 so Sn=\[\frac{ a }{ 1-r }\] so here a=1/3 and r=1/3
on substituting, you will yield the result
as n tends to infinity u can use the formula for Sum to Infinite G.P as Mentioned by Joseph
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okay a= 1/3 r=?
r = common ratio or difference
and that is -2/9
Sorry it is common ratio since it is a GP
r = t2/t1
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t2 = second term t1 = 1st term
srry 9/3=3
1/3
Yup
@Joseph91 & @Yahoo! thank u
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