Mathematics
11 Online
OpenStudy (rainbow_dash):
Find S(n) for the given arithmetic series.
a = 18, d = –10, n = 12
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OpenStudy (lgbasallote):
you do know you can just edit questions right?
OpenStudy (rainbow_dash):
I did not know that. I am noob of openstudy
OpenStudy (lgbasallote):
anyway.. find a_n first using \[\huge a_n = a_1 + (n-1)d\]
OpenStudy (rainbow_dash):
okay one sec
OpenStudy (lgbasallote):
go ahead
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OpenStudy (rainbow_dash):
-92
OpenStudy (anonymous):
isn't there direct formula for S_n ?
OpenStudy (lgbasallote):
\[\huge S_n = \frac{n(a_1 + a_n)}{2}\]
OpenStudy (lgbasallote):
you need to find a_n first
OpenStudy (rainbow_dash):
yeah^that
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OpenStudy (anonymous):
\(S_n=n/2(2a+(n-1)d\)
OpenStudy (lgbasallote):
also..how did you get -92?
OpenStudy (anonymous):
\(S_n=n/2(2a+(n-1)d)\)
OpenStudy (rainbow_dash):
well your equation. 18+(11)-10, -10*11=-110+18=-92.
OpenStudy (lgbasallote):
nevermind..it's right
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OpenStudy (lgbasallote):
now use that very long formula
OpenStudy (anonymous):
@Rainbow_Dash
this tutorial can be helpful to you on this topic.
OpenStudy (rainbow_dash):
thanks gohan
OpenStudy (anonymous):
welcome :)
go through that tutorial.
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OpenStudy (rainbow_dash):
will do.
wait, @lgbasallote , do i use your second formula with S(n)= n(a1+a_n)/2 or gohangokus?
OpenStudy (lgbasallote):
they are the same
OpenStudy (anonymous):
since u found an,u use lg's formula
OpenStudy (rainbow_dash):
awesome sauce. thanks guys!
OpenStudy (lgbasallote):
\[\LARGE S_n = \frac{n(a_1 + a_n)}{2} \implies \frac n2 (a_1 + a_1 + (n-1)d)\]\[\LARGE \implies \frac n2 (2a_1 + (n-1)d)\]
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OpenStudy (lgbasallote):
it's just a fancier (and longer) way of writing it