Find S11 for 3 + 1 + (-1) + (-3) +…
A. -77
B. -75
C. -73
D. -71

- anonymous

Find S11 for 3 + 1 + (-1) + (-3) +…
A. -77
B. -75
C. -73
D. -71

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- jdoe0001

hmm doesn't look like a sequence to me for one

- jdoe0001

can you post a quick screenshot of the material? so we can see what you mean

- Astrophysics

Nope, it's a series

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## More answers

- jdoe0001

hmm ok

- Astrophysics

So we can use sum of an arithmetic series equation I guess which is \[S_n = \frac{ n }{ 2 }[2a+(n-a)d]\] a = first term in series, d = common difference

- Astrophysics

n is what you're looking for in this case n = 11

- Astrophysics

So what's the common difference, look at your previous post what I wrote when you left

- Astrophysics

So I'll post it again for reference, the common difference is \[t_2-t_1=t_3-t_2\]

- anonymous

I'm really confused right now..

- Astrophysics

I would say so you jumped from a sequence from your last question to a series here, so just focus on the equation for now. The t here represents the term and the subscript is the address of the term, so \[t_1 = 3~~~t_2 = 1\] etc

- welshfella

Common difference can be worked out by subtracting the first term from the second

- Astrophysics

\[t_2-t_1 = t_3-t_2 \implies 1-3 = -1-1 \implies -2 = -2 \checkmark\] so it works out, our common difference d = -2

- Astrophysics

With me so far?

- anonymous

yes!

- anonymous

i'm just writing everything down in my notebook!

- Astrophysics

Ok cool :)
Well now lets find a, what is a?

- anonymous

is already one of the terms given, or do I have to find it?

- Astrophysics

a = first term in series

- anonymous

A not already sorry

- anonymous

so its 3

- Astrophysics

Good!

- Astrophysics

So we now have d = -2, a = 3, n = 11

- Astrophysics

Now we can use the equation I mentioned above \[S_n = \frac{ n }{ 2 }[2a+(n-a)d]\] can you plug it all in now

- anonymous

sn=11/2 (2(3)+(11+3)-2) correct?

- Astrophysics

Close, your sign in (n-a) is off but it looks good \[S_{11} = \frac{ 11 }{ 2 }[2(3)+(11-3)(-2)]\]

- anonymous

i got -55 ?

- Astrophysics

Yup, that looks good

- anonymous

but it isn't one of the choices?

- Astrophysics

Mhm...

- Astrophysics

So is your question asking for \[S_{11}\]

- anonymous

yes!

- Astrophysics

So it's exactly Find S11 for 3 + 1 + (-1) + (-3) +…
right

- anonymous

yes!

- Astrophysics

Mhm, everything looks right

- Astrophysics

Ok lets see...a = 3, d = -2, n = 11

- Astrophysics

I have located the problem it's the original equation I gave you, it should (n-1) not (n-a) sorry! \[S_n = \frac{ n }{ 2 }[2a+(n-1)d] = \frac{ 11 }{ 2 }[2(3)+(11-1)(-2)] = \]

- Astrophysics

Try now :)

- Astrophysics

So the equation is \[S_n = \frac{ n }{ 2 }[2a+(n-\color{red}1)d]\]

- anonymous

I got 72.8? so -73?

- Astrophysics

Nope, you must've put in wrong

- Astrophysics

\[\frac{ 11 }{ 2 }[2(3)+(11-1)(-2)] = \frac{ 11 }{ 2 }[6+10(-2)] = \frac{ 11 }{ 2 }[6-20]=\frac{ 11 }{ 2 }(-14) = -77\]

- Astrophysics

Use your order of operations :-)

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