Home > Sequences and Series > Geometric Series

## Geometric Series

A geometric series is one where to get from one term to the next you ALWAYS multiply by a constant value.

For example consider $3+6+12+24+48+...$  This is a geometric series because to get from one term to the next you always multiply by 2.  Therefore there is a common ratio.

Definitions:

a is the first term

r is the common ratio

n is the term position (or the term name)

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

$S_n$ is the sum of the first n terms

The formulae

 $S_n$ $a$ $ar$ $ar^2$ … $ar^{n-3}$ $ar^{n-2}$ $ar^{n-1}$ $rS_n$ $ar$ $ar^2$ $ar^3$ … $ar^{n-2}$ $ar^{n-1}$ $ar^n$ $S_n - rS_n$ $a$ $-ar^n$

So $S_n - rS_n=a-ar^n$

factorising each side

$\Rightarrow S_n(1-r)=a(1-r^n)$

$\Rightarrow S_n=\frac{a(1-r^n)}{1-r}$

Also it can be seen that the $n^{th}$ term $=ar^{n-1}$

Examples

Lantarna wants to save up to buy a car.  She decides to put some money aside each week and buy whatever she can after one year.  In the first week she saves $20 and each week after that she saves 5% more than the previous week. How much will she save in week 52? How much will she have saved in total after 1 year? $a =20$, because that is the first value $d=1.05$, because that is 105% which is a 5% increase $n=52$, because we are considering 52 weeks and 1 year $n^{th}=ar^{n-1} \Rightarrow 52^{nd}=20\times 1.05^51=$240.82$

$S_n=\frac{a(1-r^n)}{1-r} \Rightarrow S_{52}=\frac{20(1-1.05^{52})}{1-1.05}=4657.12$

Niks is saving up to buy one of those ugly bald cats,  It costs $5000. She plans to save$10 in the first week and each week after that she hopes to save 4% more than the week before.  How many weeks will it take her to have enough money to by the cat?

$a=10$, the amount saved in the first week

$r=1.04$, since 1.04 is the multiplier for a 4% increase

$S_n=5000$, because that is the total amount she needs to save

Since $S_n=\frac{a(1-r^n)}{1-r} \Rightarrow 5000=\frac{10(1-1.04^n)}{1-1.04}$

$\Rightarrow -200=10(1-1.04^n)\Rightarrow -20=1-1.04^n \Rightarrow 1.04^n=21$

To solve equations of this form we need to use logs.

$log(1.04^n)=log 21$

$nlog 1.04=log 21$

$n=\frac{log 21}{log 1.04}=77.6$

So after 77 weeks she will not have quite enough, but after 78 weeks she can buy the cat.

Ottilie sees two advertisements for pension schemes that both start to at the age of 65.

Scheme A:  Starts buy paying $4000 a month and each month it pays 2% less than the one before. Scheme B: Pays$2250 a month every month until you die.

After how long will it be before Ottilie is earning more per month from scheme B.

(We need to find out when they are paying the same amount and then we know that any time after that scheme B is better.)

Scheme A

$a=4000, r=0.98\Rightarrow n^{th}=4000\times 0.98^{n-1}$

Scheme B

$n^{th}=2250$

So $4000\times 0.98^{n-1}=2250 \Rightarrow 0.98^{n-1}=0.5625\Rightarrow log (0.98^{n-1})=log 0.5625$

$\Rightarrow (n-1)log 0.98=log 0.5625 \Rightarrow n=\frac{log 0.5625}{log 0.98}+1=29.5$

So scheme B will start paying more after 30 months

Anna’s parents send her away to do her sixth form studies.  They agree to pay her a gradually increasing allowance.  The allowance will go up by the same amount each month.

In the $5^{th}$ month she got $1850. In the $8^{th}$ month she got$1910.

What will she get in the $24^{th}$ month and how much will her parents have given her over the two year in total?

First consider $n=5$

So $5^{th}=1850\Rightarrow 1850=ar^4\rightarrow 1$

now consider $n=8$

So $8^{th}=1910\Rightarrow 1910=ar^7\rightarrow 2$

Divide equation 2 by equation 1

$r^3=\frac{1910}{1850}\Rightarrow r=1.011\Rightarrow a=1772.92$

Now let $n=24$

$24^{th}=1772.92\times 1.011^23=2264.45$

$S_{24}=\frac{1772.92(1-1.011^24)}{1-1.011}=48218.56$

So in the $24^{th}$ month Anna will get $2264.45 Anna will get$48218.56 in total.

### Summing to infinity

Given that $S_n=\frac{a(1-r^n)}{1-r}$

consider what happens wheh $n\rightarrow \infty$

If $-1 then $r^n\rightarrow 0$

So $S_{\infty}=\frac{a}{1-r}$

Example

Alex wants to go to a Cliff Richard concert (her all time favourite singer, in fact, she is his biggest fan).  The ticket costs $160 and her parents agree to buy it for her provided she pays them back. She asks if she can give them$15 the first week and then 10% less each week after that forever.  Does she save money and if so by how much?

$a=15, r=0.9,n=\infty$

So $S_{\infty}=\frac{15}{1-0.9}=150$

So Alex does save money.  In fact she saves \$10.  However, after going to the concert she decides that maybe she would be better off following Alvin and the Chipmunks.