anonymous
  • anonymous
Find S11 for 3 + 1 + (-1) + (-3) +… A. -77 B. -75 C. -73 D. -71
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
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jdoe0001
  • jdoe0001
hmm doesn't look like a sequence to me for one
jdoe0001
  • jdoe0001
can you post a quick screenshot of the material? so we can see what you mean
Astrophysics
  • Astrophysics
Nope, it's a series

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jdoe0001
  • jdoe0001
hmm ok
Astrophysics
  • 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
  • Astrophysics
n is what you're looking for in this case n = 11
Astrophysics
  • Astrophysics
So what's the common difference, look at your previous post what I wrote when you left
Astrophysics
  • Astrophysics
So I'll post it again for reference, the common difference is \[t_2-t_1=t_3-t_2\]
anonymous
  • anonymous
I'm really confused right now..
Astrophysics
  • 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
  • welshfella
Common difference can be worked out by subtracting the first term from the second
Astrophysics
  • 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
  • Astrophysics
With me so far?
anonymous
  • anonymous
yes!
anonymous
  • anonymous
i'm just writing everything down in my notebook!
Astrophysics
  • Astrophysics
Ok cool :) Well now lets find a, what is a?
anonymous
  • anonymous
is already one of the terms given, or do I have to find it?
Astrophysics
  • Astrophysics
a = first term in series
anonymous
  • anonymous
A not already sorry
anonymous
  • anonymous
so its 3
Astrophysics
  • Astrophysics
Good!
Astrophysics
  • Astrophysics
So we now have d = -2, a = 3, n = 11
Astrophysics
  • 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
  • anonymous
sn=11/2 (2(3)+(11+3)-2) correct?
Astrophysics
  • Astrophysics
Close, your sign in (n-a) is off but it looks good \[S_{11} = \frac{ 11 }{ 2 }[2(3)+(11-3)(-2)]\]
anonymous
  • anonymous
i got -55 ?
Astrophysics
  • Astrophysics
Yup, that looks good
anonymous
  • anonymous
but it isn't one of the choices?
Astrophysics
  • Astrophysics
Mhm...
Astrophysics
  • Astrophysics
So is your question asking for \[S_{11}\]
anonymous
  • anonymous
yes!
Astrophysics
  • Astrophysics
So it's exactly Find S11 for 3 + 1 + (-1) + (-3) +… right
anonymous
  • anonymous
yes!
Astrophysics
  • Astrophysics
Mhm, everything looks right
Astrophysics
  • Astrophysics
Ok lets see...a = 3, d = -2, n = 11
Astrophysics
  • 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
  • Astrophysics
Try now :)
Astrophysics
  • Astrophysics
So the equation is \[S_n = \frac{ n }{ 2 }[2a+(n-\color{red}1)d]\]
anonymous
  • anonymous
I got 72.8? so -73?
Astrophysics
  • Astrophysics
Nope, you must've put in wrong
Astrophysics
  • 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
  • Astrophysics
Use your order of operations :-)

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