Question

What is the sum of a 17-term arithmetic sequence where the first term is 6 and the last term is -90?

Answer 1

n(a1+an)/2

Answer 2

Use the definition of an arithmetic sequence, and the required algebra to relate the first and last terms.

Answer 3

idk how to do this at all

Answer 4

what do i plug into what

Answer 5

here..n=no.of terms.
a1=first term.
an=last term.

Answer 6

First things first, @helpisuck
What IS an arithmetic sequence (progression) ?

U

Answer 7

Or, on second thought, since you're just looking for the sum, anyway, @dlearner 's got it down...
my bad :3

Answer 8

so (17+6)/2?

Answer 9

what do i do with the -90?

Answer 10

this website is a good high school prep review for the Regents, a NY mandate test for all HS to take before they graduate.
http://www.regentsprep.org/Regents/math/algtrig/ATP2/ArithSeq.htm

Answer 11

ima have to look at that later

Answer 12

you have to look at it now since the answers to the questions (except a direct answer to your problem) you have at the moment are in that link

Answer 13

No, @helpisuck \[\Large \frac{n(a_1 + a_n)}{2}\]
is the sum of the first n-terms of an arithmetic progression, true enough, but your 'plugging-in' is faulty...

Now, let's replace the values one at a time, shall we?
n here represents the number of terms you're adding, so what is n?

Answer 14

i understand all that im just confused on how to plug it in. im not good with over reading on math i need it pretty much spelled out for me then explained why

Answer 15

we'll make u spell it out :)

Answer 16

say again lol

Answer 17

basically

Answer 18

i will have to but i cant now

Answer 19

\[\sum_{k =1}^{n}k=\frac{ n(n + 1) }{ 2}\]
negative the sum to 90 plus the sum to 6

Answer 20

@helpisuck An arithmetic sequence is given in the form:\[\bf \left\{ a_n \right\} =\left\{ a,a+d,a+2d,a+3d...,a+(n-1)d \right\}\]We are given that the first term is 6 hence \(\bf a = 6\). We are also given the last term of this 17-TERM sequence as -90 hence \(\bf a+(17-1)d = 6+16d=-90\). Solving for 'd' gives us:
\[\bf 6+16d=-90 \implies d = \frac{ -90-6 }{ 16 }=-6\]Now we have everything we need to compute the sum of the 17 terms of this sequence. The sum of any arithmetic sequence is given by:\[\bf S_n=\frac{ n }{ 2 }(2a+(n-1)d) \implies S_n=\frac{ n }{ 2 }(12+(n-1)(-6))\]Now simply evaluate \(\bf S_n\) when \(\bf n=17\).

Can you do that?

Answer 21

ive never seen the formula before either of them ive seen the first formlua that was posted but idk what to plug in where?

Answer 22

can i ask you another question? if i have a 55 in my class and my final exam is coming up and it is worth like 320 points what would i need on the test to get a passing grade? (75)

Answer 23

We are getting the sum of the first 17 terms. The formula for the first 'n' terms of an arithmetic sequence is given by \(\bf S_n=\frac{n}{2}(2a+(n-1)d)\) where \(\bf S_n\) represents the sum of the first \(\bf n\) terms. We figured from the given information that the first term is 6, i.e. \(\bf a=6\). We were also given that the last term, which is the 17th term is -90, i.e.
\(\bf a+(17-1)d=6+16d=-90\). From here I solved for 'd' which gave me \(\bf d=-6\). Now we use the formula for \(\bf S_n\) which is the sum of the first n terms of an arithmetic sequence. Since we know what 'a' and 'd' are, we plug that in to our \(\bf S_n\) formula and we get:\[\bf S_n=\frac{ n }{ 2}(2(6)+(n-1)(-6))=\frac{ n }{ 2 }(12+(n-1)(-6))\]Now since we are finding the sum of the first 17 terms, we let \(\bf n = 16\) and evaluate \(\bf S_n\) accordingly. Can you do that now?

@helpisuck

Answer 24

Two things @helpisuck
1: We can't answer that ^ unless we know how 'heavy' exactly your final exam is...
2: Just do your best on that exam. Can't go wrong that way. If you can perfect it (I have faith in you ^_^) then perfect it...

And that starts with understanding concepts ^

Don't worry if you think you're being spoonfed, we were all spoonfed before we learned to pick up our own cutlery :P

Answer 25

@helpisuck First finish up this question and then make a new question..

Answer 26

It depend on your performance prior to final and whether grades are on the standard bell curve, only so many A, B, C, D, F.

Answer 27

@helpIsuck has left...

Maybe it wasn't such a good idea to be so technical so early on @genius12 :3

Answer 28

@terenzreignz Lol I was trying to make as much sense of the knowledge as I possibly could. I also assumed that he had some understanding of such sequences...