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## Arithmetic Series

An arithmetic seires is one in which to get from one term to the next you add a constant value.

For example consider the series $3+7+11+15+19+...$ To get from one term to the next her you ALWAYS add 4.

Definitions

a is the value of the first term

d is the common diference (the amount we add each time)

n is a term position

$n^{th}$ is the value at position n

$S_n$ is total of the first n terms

Formulae

 $S_n$ a a+d a+2d … a+(n-3)d a+(n-2)d a+(n-1)d $S_n$ a+(n-1)d a+(n-2)d a+(n-3)d … a+2d a+d a $2S_n$ 2a+(n-1)d 2a+(n-1)d 2a+(n-1)d … 2a+(n-1)d 2a+(n-1)d 2a+(n-1)d

The first row is the series written smallest first

The second row is the series in reverse

The third row is the top two rows added together.  Notice that in the third row we get the same term each time and that the total of this new series is going to be twice the original.

The $n^{th}$ term with be the term in position

$\therefore n^{th}=a+(n-1)d$

$S_n$ will be half the value of the total of the third row

$\therefore S_n=\frac{n}{2}(2a+(n-1)d)$

Examples

A series starts with a value of 4 and to get to the next term you add 11 each time.

What is the value of the $20^{th}$ term?

What is the total of the first 20 terms?

$a=4$

$d=11$

Let $n=20$, since we are interested in the first 20 terms.

$\therefore n^{th}=a+(n-1)d$

$20^{th}=4+(20-1)\times 11=213$

$S_n=\frac{n}{2}(2a+(n-1)d)$

$S_{20}=\frac{20}{2}(2\times 4+(20-1)\times 11)=2170$

A series starts with 8 and the sum of the first 16 terms is 584.

Given that the series is arithmetic, goes up by the same constant value each time:

What is the commom difference?

What is the sum of the first 20 terms?

$a=8$, since this is the first term.

$S_{16}=584$, since 584 is the total of the first 16 terms.

Let $n=16$, because we are looking at the first 16 terms.

$S_n=\frac{n}{2}(2a+(n-1)d)$

$\Rightarrow 584=\frac{16}{2}(2\times 8 + (16-1)d)$

$\Rightarrow 584=8(16 + 15d)=128+120d$

$\therefore d = 3.8$

An arithmetic series has a common difference of 5 and the $60^{th}$ term 435, what is the sum of the first 60 terms?

$d = 5$, it is the common difference.

$60^{th}=435$, given.

Let n = 60, because we are looking at the $60^{th}$ term and the sum of the first 60 terms.

$n^{th}=a+(n-1)d\Rightarrow 435=a+59\times 5\Rightarrow a = 140$

$S_n=\frac{n}{2}(2a+(n-1)d)$

So, $S_{60}=\frac{60}{2}(280+59\times 5)$

$\therefore S_{60}=17250$

An arithmetic series has a $5^{th}$ term of 18 and a $12^{th}$ term of 46.

What is the sum of the first 50 terms?

For this problem we need to consider when $n=5$ and when $n=12$

(Note: $n^{th}=a+(n-1)d$)

 When $n=5$$5^{th}$=18 $\therefore 18=a+4d$ When $n=12$$12^{th}$=46 $\therefore 46=a+11d$

Solving the two resulting simultaneous equations gives $a=2,d=4$

Now let $n=50$, because we want the sum of the first 50 terms.

$S_{50}=\frac{50}{2}(4+49\times 4)=5000$

The sum of the first 10 terms of an arithmetic series is 120.

The sum of the first 16 terms is 80.

What is the size of the first term?

$S_n=\frac{n}{2}(2a+(n-1)d)$

When $n=10$

$120=\frac{10}{2}(2a+9d)\Rightarrow 120=10a+45d$

When $n=16$

$80=\frac{16}{2}(2a+15d)\Rightarrow 80=16a+120d$

Solving the simultaneous equations above gives $a=\frac{45}{2}, d=-\frac{7}{3}$

Shannon is saving up to buy a pony.  In the first week she saves $10, in the second she saves$12.  She foolishly believes that she and keep on saving $2 more than she did in the previous week. The pony costs$10 000.  How long will it take her to save up to buy the pony if she manages to save at her expected rate.

Here we do not know n.

$a=10$, $d=2$ and $S_n=10 000$.

We require the total amount saved so we shall use $S_n=\frac{n}{2}(2a+(n-1)d)$

$10 000=\frac{n}{2}(20+2(n-1))$

$\Rightarrow n^2+9n-10000=0$

$\Rightarrow n=95.6$ or $n=-104.6$

Clearly $n>0\Rightarrow n=95.6$

It follows that after 95 weeks she will not have quite enough money, so it will take 96 weeks.