find the sum of the given arithmetic series. (: PLEASE HELP ME!!! 1.30+23+16+9+...(-30) 2. 26+19+12+5+...+(-37) 3.27+20+13+6+...+(-36)
just use a caclulator lol
and it will give me the answers? lol
dont listen to @mitu12 Mathway is a site that gives the answers and doesnt explain. it does but you have to pay and sometimes you dont understand the concept.
ohhh okay thank you!
its like a calculator
The sum of an arithmetic series is given by:\[\Large\rm S_n=\frac{n(a_1+a_n)}{2}\]
We're given the last term and the first term. The work we'll have to do is figuring out `how many` terms we have. That is our \(\Large\rm n\) value.
The first is \(\Large\rm a_1=30\) The common difference is \(\Large\rm d=a_2-a_1=-7\) The general form for finding the nth term of a geometric sequence is given by:\[\Large\rm a_n=a_1+d(n-1)\]Plugging in our stuff gives us:\[\Large\rm a_n=30-7(n-1)\]
We know that our LAST term is -30. Let's plug that in for our \(\Large\rm a_n\) and see which \(\Large\rm n\) value that corresponds to. That will tell us how many terms we have.
\[\Large\rm -30=30-7(n-1)\]
Do you understand how to solve for n in that set up?
do i just try and get n by itself?
Yes, then we will have an \(\Large\rm n\) value for our \(\Large\rm S\) formula :)
okay!
Hmm it's not working out to an integer value.. this is really weird :o
Oh I see the problem. The first one listed is NOT an arithmetic series. One of the numbers is messed up. We cant have 30 AND -30 in the series. Did you copy it down correctly? I assume you just copy/pasted, yes? :o
sorry i meant (-33)
Oh ok that would make more sense :3 So we're actually plugging it in like this:\[\Large\rm -33=30-7(n-1)\]
okay sorry about that again!!
-55n?
Hmmm, I'm not sure what you did there. We're trying to isolate our n.\[\Large\rm -33=30-7(n-1)\]So start by subtracting 30 from each side,\[\Large\rm -63=-7(n-1)\]Divide by -7,\[\Large\rm 9=n-1\]Add 1 to each side,\[\Large\rm 10=n\]Understand those steps ok? :o
yes. lol
So now we have all of the information we need: \(\Large\rm a_1=30, \qquad a_n=-33,\qquad n=10\) Plug them into your formula for sum,\[\Large\rm S=\frac{n(a_1+a_n)}{2}\]What do you get for your sum? :)
-15
do you mind helping me with the other two? it is okay if you do not want too.
So you'll need to do these steps for all of them, so get comfortable with the process. \[\Large\rm S=\frac{n(a_1+a_n)}{2}\]We need three things, \(\Large\rm a_1,~ a_n,~n\) We already have \(\Large\rm a_1,~a_n\) In order to find \(\Large\rm n\), we have to first find the common difference and plug everything into our formula for the nth term.\[\Large\rm a_n=a_1+d(n-1)\]
So start by finding your common difference. (The second term minus the first term). Then plug all that jazz into the formula to find your n.
okay thank you!!
You can do it mads! c: Lemme know if you get stuck. Look back at the steps we took also.
was the -15 correct? lol
For the first one? Yes, good job c:
Thank you for all your help(:
i got -11 for the 2nd one and -9 for the 3rd. Do you mind checking to see if i got the correct answer? :)
For the second problem, \[\Large\rm a_1+a_n=-11\]Did you forgot to multiply that by \(\Large\rm \dfrac{n}{2}\)? Remember,\[\Large\rm S=\frac{n}{2}(a_1+a_n)\]
You have to find your n value :O
ive confused myself now i am lost..im so sorry.
i got,-10.57 but i think i did something wrong on my work.
What is the common difference? \(\Large\rm d=?\)
-7
\[\Large\rm a_n=a_1+d(n-1)\]\[\Large\rm -37=26-7(n-1)\]And what'd you get for n? :d
-10.57?
Hmm, no. Make sure you do the subtraction step before you do the division step.
okay, i see what i did wrong...
That's only half of the work though! Remember you have to take your n value and plug it into the S formula.
nevermind i messed up again. i am so sorry.
lol :) dat mads
|dw:1406691200173:dw|Is this step ok?
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