## anonymous one year ago Find S11 for 3 + 1 + (-1) + (-3) +… A. -77 B. -75 C. -73 D. -71

1. jdoe0001

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

2. jdoe0001

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

3. Astrophysics

Nope, it's a series

4. jdoe0001

hmm ok

5. 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

6. Astrophysics

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

7. Astrophysics

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

8. Astrophysics

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

9. anonymous

I'm really confused right now..

10. 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

11. welshfella

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

12. 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

13. Astrophysics

With me so far?

14. anonymous

yes!

15. anonymous

i'm just writing everything down in my notebook!

16. Astrophysics

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

17. anonymous

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

18. Astrophysics

a = first term in series

19. anonymous

20. anonymous

so its 3

21. Astrophysics

Good!

22. Astrophysics

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

23. 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

24. anonymous

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

25. Astrophysics

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

26. anonymous

i got -55 ?

27. Astrophysics

Yup, that looks good

28. anonymous

but it isn't one of the choices?

29. Astrophysics

Mhm...

30. Astrophysics

So is your question asking for $S_{11}$

31. anonymous

yes!

32. Astrophysics

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

33. anonymous

yes!

34. Astrophysics

Mhm, everything looks right

35. Astrophysics

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

36. Astrophysics

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

37. Astrophysics

Try now :)

38. Astrophysics

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

39. anonymous

I got 72.8? so -73?

40. Astrophysics

Nope, you must've put in wrong

41. Astrophysics

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

42. Astrophysics

Use your order of operations :-)