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Mathematics 11 Online
OpenStudy (rainbow_dash):

Find S(n) for the given arithmetic series. a = 18, d = –10, n = 12

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

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

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

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.

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):

\[\LARGE S_n = \frac{n(a_1 + a_n)}{2} \implies \frac n2 (a_1 + a_1 + (n-1)d)\]\[\LARGE \implies \frac n2 (2a_1 + (n-1)d)\]

OpenStudy (lgbasallote):

it's just a fancier (and longer) way of writing it

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