Question

Find S11 for 3 + 1 + (-1) + (-3) +…

A. -77
B. -75
C. -73
D. -71

Answer 1

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

Answer 2

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

Answer 3

Nope, it's a series

Answer 4

hmm ok

Answer 5

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

Answer 6

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

Answer 7

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

Answer 8

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

Answer 9

I'm really confused right now..

Answer 10

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

Answer 11

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

Answer 12

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

Answer 13

With me so far?

Answer 14

yes!

Answer 15

i'm just writing everything down in my notebook!

Answer 16

Ok cool :)

Well now lets find a, what is a?

Answer 17

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

Answer 18

a = first term in series

Answer 19

A not already sorry

Answer 20

so its 3

Answer 21

Good!

Answer 22

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

Answer 23

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

Answer 24

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

Answer 25

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

Answer 26

i got -55 ?

Answer 27

Yup, that looks good

Answer 28

but it isn't one of the choices?

Answer 29

Mhm...

Answer 30

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

Answer 31

yes!

Answer 32

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

Answer 33

yes!

Answer 34

Mhm, everything looks right

Answer 35

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

Answer 36

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

Answer 37

Try now :)

Answer 38

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

Answer 39

I got 72.8? so -73?

Answer 40

Nope, you must've put in wrong

Answer 41

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

Answer 42

Use your order of operations :-)