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OpenStudy (kanwal32):
(2/3)^n-1
OpenStudy (kanwal32):
to infinite or 4
OpenStudy (kanwal32):
no i am asking till where we have to find infinite or 4
OpenStudy (anonymous):
@kanwal32
OpenStudy (anonymous):
you mean this?
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OpenStudy (sdfgsdfgs):
yes, thats what he is asking. But do u understand the solution given in the old post? u can simply substitute 3/2 by 2/3 to get ur ans...
OpenStudy (sdfgsdfgs):
there is a formula given in the old post on how to calculate the sum....did u see it?
OpenStudy (anonymous):
I did, but it's pretty confusing, could you start me out?
OpenStudy (sdfgsdfgs):
u need to understand the 3 terms used in the formula: a, R and n
a is the initial term in the series so in this case, it is equal to
\[(\frac{ 2 }{ 3})^{1-1} \]
=1
ok?
OpenStudy (anonymous):
Okay.
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OpenStudy (sdfgsdfgs):
R is the common ratio...so what is the ratio between the first and second term in this series?
Basically what u get if u divide the second term by the first term? u already know the first term, a, is 1...
OpenStudy (sdfgsdfgs):
nope, first term is 1....what is the second term - when n=2....