Question

What is the sum of a 30-term arithmetic sequence where the first term is 72 and the last term is -102?

Answer 1

do you know the equation for this?

Answer 2

An=a1+(n-1)d

Answer 3

\[
a_n = a_1 + (n-1)d \\
a_{30} = 72 + (30-1)d
\]Solve for \(d\) first to get the sequence formula.

Answer 4

so you would do what @wio said

Answer 5

\[
a_{30}=-102 = 72+(30-1)d
\]

Answer 6

so d would equal -6?

Answer 7

\[\begin{split}
S_k
&= \sum_{n=1}^{k}a_n \\
&= \sum_{n=1}^{k}a_1+(n-1)d \\
&= \sum_{n=1}^{k}a_1+nd-d \\
&= \sum_{n=1}^{k}a_1+\sum_{n=1}^{k}nd-\sum_{n=1}^{k}d \\
&= ka_1+d\sum_{n=1}^{k}n-kd \\
&= ka_1+d\frac{k(k+1)}{2}-kd \\
&=k\left(a_1 + d\frac{1-k}{2}\right)
\end{split}\]
This is my attempt to calculate the partial sum, given \(k\) terms.

Answer 8

@soccerbabe239 that looks correct.

Answer 9

that equation looks soo confusing... explain it a little?

Answer 10

Well \[\sum\]Just means we're adding up each term in the sequence.

Answer 11

It's well know that \[
\sum^{k}_{n=1} n = \frac{k(k+1)}{2}\quad\text{and}\quad \sum^{k}_{n=1} 1 = k
\]

Answer 12

Don't worry about it though

Answer 13

They should have given you a formula for computing partial sums.

Answer 14

I didn't see one but I think ill just do it the long way:) by figuring out all 30 numbers then adding them all together

Answer 15

Well, first try using the formula I gave you.

Answer 16

I got -450 as my answer

Answer 17

http://www.wolframalpha.com/input/?i=sum+1+to+30%2C+72+%2B+%28n-1%29%28-6%29

Says it is -450

Answer 18

\[Sn \frac{ n(A1+An) }{ 2 }\] is the formula I was given in my lesson I just looked

Answer 19

thank you by the way:)

Answer 20

Hmm, then use that one, it looks legit.

Answer 21

haha yeah it does

Answer 22

I made a small error on my formula, it should be: \[
\begin{split}
S_k
&= \sum_{n=1}^{k}a_n \\
&= \sum_{n=1}^{k}a_1+(n-1)d \\
&= \sum_{n=1}^{k}a_1+nd-d \\
&= \sum_{n=1}^{k}a_1+\sum_{n=1}^{k}nd-\sum_{n=1}^{k}d \\
&= ka_1+d\sum_{n=1}^{k}n-kd \\
&= ka_1+d\frac{k(k+1)}{2}-kd \\
&=k\left(a_1 + d\frac{k-1}{2}\right) \\
&=k\left(\frac{2a_1}{2} + \frac{(k-1)d}{2}\right) \\
&=k\left( \frac{a_1+a_1+(k-1)d}{2}\right) \\
&=k\left( \frac{a_1+a_k}{2}\right) \\
\end{split}
\]Which ends up being the same formula.

Answer 23

Interestingly enough, it's just the average between the first and last term, multiplied by the number of terms. It makes a lot of sense.

Answer 24

The sum for any arithmetic series is given by the formula:\[S _{n}=\frac{ n }{ 2 }(2a+(n-1)d)\]I can prove it using a really neat trick which doesn't require partial sums but I think we will skip the proof here. Although the partial sum proof from @wio was correct although you made a mistake when you factored k out.

Anyway, the sum of the first 30 terms then is:\[S _{30 }=\frac{ 30 }{ 2 }(2(72)+(30-1)(-6))=15(144+(29)(-6))=15(-30)=-450\]@soccerbabe239

Answer 25

@soccerbabe239

Answer 26

thank you very much! that helps a lot haha!