www.mathcentre.ac.uk/.../Arithmetic%20and%20geometric%20progressions.mp4

In this video, we're going to be
looking at sequences and series,

so let's begin by looking at
what a sequences.

This, for instance is a

sequence. It's a
set of numbers.

And here we seem to have a rule.
All of these are odd numbers, or

we can look at it. We increase
by two each time 13579.

So there's our sequence of
odd numbers.

Here is another sequence.

These numbers
are the

square numbers.
1 squared, 2 squared, is 4 three

squared is 9. Four squared is 16
and 5 squared is 25. So again

we've got a sequence of numbers.
We've got a rule that seems to

produce them. Those are the.

Square numbers. Is a
slightly different sequence.

Here we've got alternation
between one and minus one back

to one on again to minus one
back to one on again to minus

one. But this is still a
sequence of numbers.

Now, because I've written some
dots after it here.

This means that this is meant to
be an infinite sequence. It goes

on forever and this is meant to
be an infinite sequence. It

carries on forever, and this one
does too. If I want a finite

sequence. What might
a finite sequence

look like, for

instance 1359? That would
be a finite sequence. We've got

4 numbers and then it stops

dead. Perhaps if we look
at the sequence of square

numbers 1, four, 916, that again
is a finite sequence.

Sequence that will be very
interested in is the sequence

of whole numbers, the counting
numbers, the integers. So

there's a sequence of integers,
and it's finite because it

stops at N, so we're
counting 123456789 up to N.

And the length of this sequence
is an the integer, the number N.

Very popular sequence of
numbers. Quite well known is

this particular sequence.

This is a slightly different
sequence. It's infinite keeps on

going and it's called the

Fibonacci sequence. And we can
see how it's generated. This

number 2 is formed by adding the
one and the one together, and

then the three is formed by
adding the one and the two

together. The Five is formed by
adding the two and the three

together, and so on.

So there's a question here. How
could we write this rule down in

general, where we can say it
that any particular term is

generated by adding the two
numbers that come before it in

the sequence together? But how
might we set that down? How

might we label it? One way might
be to use algebra and say will

call the first term you, and
because it's the first time we

want to label it, so we call it

you one. And then the next
terminal sequence, the second

term. It would make sense to
call it you too, and the third

term in our sequence. It would
make sense therefore, to call it

you 3. You four and so on
up to UN. So this represents a

finite sequence that's got N
terms in it. If we look at

the Fibonacci sequence as an
example of making use of this

kind of notation, we could say
that the end term UN was

generated by adding together the
two terms that come immediately

before it will.

Term that comes immediately
before this must have a number

attached to it. That's one less
than N and that would be N minus

one. Plus on the term that's
down, the term that comes

immediately before this one must
have a number attached to it.

That's one less than that. Well,
that's UN minus 1 - 1 taking

away 2 ones were taking away two

altogether. So that we can see
how we might use the algebra

this algebraic notation help us
write down a rule for the Fibo

Nachi sequence. OK, how can

we? Use this in a
slightly different way.

What we need to look at now
is to move on and have a

look what we mean by a series.

This is a
sequence, label it.

A sequence it's a list of
numbers generated by some

particular rule. It's finite
because there are any of them.

What then, is a series series is
what we get.

When we add.

Terms of the sequence.

Together And because
we're adding together and terms

will call this SN.

The sum of N terms
and it's that which is

the series.

So. Let's have a
look at the sequence

of numbers 123456, and
so on up to

N.

Then S1.

Is just one.

S2.

Is the sum of the
first 2 terms 1 +

2? And that gives us 3.

S3. Is
the sum of the first three

terms 1 + 2 + 3?

And that gives us 6

and S4. Is the sum
of the first four terms 1 +

2 + 3 + 4 and that

gives us. Hey.

So this gives us the basic
vocabulary to be able to move on

to the next section of the
video, but just let's remind

ourselves first of all.

A sequence.

Is a set of numbers
generated by some rule.

A series is what we get when we
add the terms of the sequence

together. This particular
sequence has N terms in it

because we've labeled each term
in the sequence with accounting

number. If you like U1U2U free,
you fall you N.

Now. With this vocabulary of
sequences and series in mind,

we're going to go on and have a
look at a 2 special kinds of

sequences. The first one is
called an arithmetic progression

and the second one is called a

geometric progression. Will
begin with an arithmetic

progression. Let's start by
having a look at this

sequence of. Odd

Numbers that we
had before 1357.

Is another sequence
not 1020 thirty,

and so on.

What we can see in this first
sequence is that each term after

the first one is formed by
adding on to.

1 + 2 gives us 3.

3 + 2 gives us 5.

5 + 2 gives us 7 and it's
because we're adding on the

same amount every time. This
is an example of what we call

an arithmetic.

Progression.
If we look at this sequence of

numbers, we can see exactly the
same property we've started with

zero. We've added on 10, and
we've added on 10 again to get

20. We've had it on 10 again to
get 30, so again, this is

exactly the same. It's an

arithmetic progression. We don't
have to add on things, so

for instance a sequence of
numbers that went like this 8

five, 2 - 1.

Minus

4. If we look what's
happening where going from 8:00

to 5:00, so that's takeaway
three were going from five to

two, so that's takeaway. Three
were going from 2 to minus. One

takeaway. Three were going from
minus one to minus four takeaway

3. Another way of thinking about
takeaway three is to say where

adding on minus three.

8 at minus three is 5 five at
minus three is 2, two AD minus

three is minus one, so again,
this is an example of an

arithmetic progression. And what
we want to be able to do is to

try and encapsulate this
arithmetic progression in some

algebra, so we'll use the letter

A. To stand for
the first term.

And will use the letter
D to stand for the

common difference. Now the
common difference is the

difference between each term and
it's called common because it is

common to each between each
term. So let's have a look at

one 357 and let's have a think
about how it's structured 13.

5. 7 and so
on. So we begin with one and

then the three is 1 + 2.

The Five is 1 + 2 tools because
by the time we got to five,

we've added four onto the one.

The Seven is one plus. Now the
time we've got to Seven, we've

added three tools on. Let's just
do one more. Let's put nine in

there and that would be 1 + 4
times by two.

So let's see if we can begin to
write this down. This is one.

Now what have we got here? This
is the second term in the

series. But we've only got 1 two
there, so if you like we've got

1 + 2 - 1 times by two.

One plus now, what's multiplying
the two here? Well, this is the

third term in the series.

So we've got a 2 here,
so we're multiplying by 3 -

1. Here this is term
#4 and we're

multiplying by three,
so that's 4 - 1 times

by two. And here this
is term #5, so we've

got 1 + 5 - 1
times by two.

Now, if we think about
what's happening here.

We're starting with A.

And then on to the A. We're

adding D. Then we're adding on
another day, so that's a plus

2D, and then we're adding on
another D. So that's a plus

3D. The question is, if we've
got N terms in our sequence,

then what's the last term? But
if we look, we can see that the

first term was just a.

The second term was a plus, one
D. The third term was a plus

2D. The fourth term was a plus
3D, so the end term must be

a plus N minus one.

Gay.

Now, this last term of
our sequence, we often label

L and call it the

last term.

Or

The end.

Turn. To be more mathematical

about it. And one of the things
that we'd like to be able to do

with a sequence of numbers like
this is get to a series. In

other words, to be able to add
them up. So let's have a look at

that. So SN the some of these
end terms is A plus A+B plus

A plus 2B plus. But I want
just to stop there and what I

want to do is I want to
start at the end. This end

now now the last one will be
plus L.

So what will be the next one
back when we generate each term

by adding on D. So we added on D
to this one to get L. So this

one's got to be L minus D.

And the one before that
one similarly will be L

minus 2D. On the rest of the
terms will be in between.

Now I'm going to use a trick.

Mathematicians often use. I'm
going to write this down the

other way around. So I have L

there. Plus L minus

D. Plus L minus 2D plus
plus. Now what will I have?

Well, writing this down either
way around, I'll Have A at the

end. Then I'll have this next

term a. Plus D.

And I'll have this next
term, A plus 2D.

Now I'm going to add these two
together. Let's look what

happens if I add SN&SN together.
I've just got two of them.

By ad A&L together I get a
plus L let me just group

those together.

Now I've got a plus D&L Minus D,
so if I add them together I have

a plus L Plus D minus D, so
all I've got left is A plus L.

But the same thing is going to
happen here. I have a plus L

Plus 2D Takeaway 2D, so again
just a plus L.

When we get down To this end,
it's still the same thing

happening. I've A plus L
takeaway 2D add onto D so again

the DS have disappeared. If you
like and I've got L plus A.

Plus a plusle takeaway D add
on DLA and right at the

end. L plus a again.

Well, how many of these have I
got? But I've got N terms.

In each of these lines of sums,
so I must still have end terms

here, and so this must be an
times a plus L.

And so if we now divide
both sides by two, we have.

SN is 1/2 of N times
by a plus L and that

gives us our some of the
terms of an arithmetic

progression. Let's just write
down again the two results that

we've got. We've got L the
end term, or the final term

is equal to a plus N
minus one times by D and

we've got the SN is 1/2.

Times by N number of
terms times by a plus

L.

Now, one thing we can do is take
this expression for L and

substitute it into here.

Replacing this al, so let's do
that. SN is equal to 1/2.

Times by N number of terms
times by a plus and instead

of L will write this a
plus N minus one times by

D. April say gives us
two way.

So the sum of the
end terms is 1/2 an

2A plus N minus 1D.

Close the bracket.

And these. That I'm
underlining are the three

important things about an
arithmetic progression.

If A is the first

term. And D is
the common difference.

And N
is the

number of
terms.

In our arithmetic progression,
then, this expression gives us

the NTH or the last term.
This expression gives us the

some of those N terms, and
this expression gives us also

the sum of the end terms.

One of the things that you also
need to understand is that

sometimes we like to shorten the
language as well as using

algebra. So that rather than
keep saying arithmetic

progression, we often refer to
these as a peas.

Now we've got some facts, some

information there. So let's have
a look at trying to see if we

can use them to solve some

questions. So let's have
a look at this

sequence of numbers again,
which we've identified.

And let's ask ourselves
what's the sum?

Of. The first

50 terms So
we could start to try and add

them up. 1 + 3 is four and four
and five is 9, and nine and

Seven is 16 and 16 and 9025, and
then the next get or getting

rather complicated. But we can
write down some facts about this

straight away. We can write down
that the first term is one.

We can write down that the
common difference Dean is 2 and

we can write down the number of
terms we're dealing with. An is

50. We know we have a
formula that says SN is 1/2

times the number of terms.

Times 2A plus N minus 1D. So
instead of having to add this up

as though it was a big
arithmetic sum a big problem, we

can simply substitute the
numbers into the formula. So SNS

50 in this case is equal to 1/2.

Times by 50.

Times by two A That's just two
2 * 1 plus N minus one

and is 50, so N minus one

is 49. Times by the common

difference too.

So. We
can cancel a 2 into the 50

that gives us 25 times by now.

2 * 49 or 2 * 49
is 98 and two is 100, so

we have 25 times by 100, so
that's 2500. So what was going

to be quite a lengthy and
difficult calculation's come out

quite quickly. Let's see if we
can solve a more difficult

problem.
1.

Plus 3.5.
+6.

Plus 8.5. Plus

Plus 101.

Add this up.

Well. Can we identify what
kind of a series this is? We can

see quite clearly that one to
3.5 while that's a gap of 2.5

and then a gap of 2.5 to 6. So
what we've got here is in fact

an arithmetic progression, and
we can see here. We've got 100

and one at the end. Our last
term is 101 and the first term

is one. Now we know a formula.

For the last term L.

Equals A plus N minus
one times by D.

Might just have a look at what
we know in this formula. What we

know L it's 101.

We know a It's the first
term, it's one.

Plus Well, we have no idea what
any is. We don't know how many

terms we've got, so that's N
minus one times by D and we know

what that is, that's 2.5.

Well, this is nothing more than
an equation for an, so let's

begin by taking one from each
side. That gives us 100 equals N

minus one times by 2.5. And now
I'm going to divide both sides

by 2.5 and that will give me 40
equals N minus one, and now I'll

add 1 to both sides and so 41 is
equal to end, so I know how many

terms that. Are in this series,
So what I can do now is I

can add it up because the sum of
N terms is 1/2.

NA plus

L. And I
now know all these terms

here have 1/2 * 41

* 1. Plus

101. Let me just turn
the page over and write this

some down again.

SN is equal
to 1/2 *

41 * 1
+ 101.

So we have 1/2 times by 41
times by 102 and we can cancel

it to there to give US 41
times by 51. And to do that

I'd want to get out my
Calculator, but we'll leave it

there to be finished.

So that's one kind of problem.

Let's have a look at another
kind of problem.

Let's say we've got an
arithmetic progression whose

first term is 3.

And the sum.

Of.

The first 8.

Terms.

Is twice.
The sum

of the
first 5

terms.

And that seems really quite

complicated. But it needn't
be, but remember this is the

same arithmetic progression.

So let's have a think what this
is telling us A is equal to

three and the sum of the first 8
terms. Well, to begin with,

let's write down what the sum of
the first 8 terms is.

Well, it's a half.

Times N Times
2A plus and

minus 1D.

And N is equal to 8.

So we've got a half.

Times 8. 2A
plus N minus one is

7D. So S 8
is equal to half of

eight is 4 * 2
A Plus 7D.

But we also know that a
is equal to three, so we

can put that in there as
well. That's 4 * 6 because

a is 3 + 7 D.

Next one, the sum
of the first 5

terms. Let me just write
down some of the first

8 terms were.

4. Times
6 minus

plus 7D
first 5

terms. Half times the number of
terms. That's 5 * 2 A plus

N minus one times by D will.
That must be 4 because any is

5 times by D.

So much is 5 over 2 and
let's remember that a is equal

to three, so that 6 + 4
D. So I've got S 8 and

I've got S5 and the question
said that S8 was equal to twice

as five. So I can write this

for S8. Is
equal to

twice. This which is
S five 2 * 5 over two

6 + 4 D and what seemed
a very difficult question as

reduced itself to an ordinary
linear equation in terms of D.

So we can do some cancelling
there and we can multiply out

the brackets for six is a 24
+ 28, D is equal to 56R.

30 + 5 fours

are 20D. I can take
20D from each side that gives me

8 D there.

And I can take 24 from each
side, giving me six there. So D

is equal to.

Dividing both sides by 8,
six over 8 or 3/4 so I know

everything now that I could
possibly want to know about

this arithmetic progression.

Now let's go on and have a look
at our second type of special

sequence, a geometric

progression. So.
Take these

two six

1854. Let's have a look
at how this sequence of numbers

is growing. We have two. Then we

have 6. And then we have
18. Well 326 and three sixes

are 18 and three eighteens are
54. So this sequence is growing

by multiplying by three each
time. What about this sequence

one? Minus

2 four. Minus

8. What's happening here? We can
see the signs are alternating,

but let's just look at the

numbers. 1 * 2 would be two 2 *
2 would be four. 2 * 4 would be

8. But if we made that minus
two, then one times minus two

would be minus 2 - 2 times minus
two would be plus 4 + 4 times by

minus two would be minus 8, so
this sequence to be generated is

being multiplied by minus two.
Each term is multiplied by minus

two to give the next term.

These are examples of geometric
progressions, or if you like,

GPS. Let's try and write one
down in general using some

algebra. So like the AP, we take
A to be the first term.

Now we need something like D.
The common difference, but what

we use is the letter R and we
call it the common ratio, and

that's the number that does the
multiplying of each term to give

the next term.

So 3 times by two gives us 6,
so that's the R. In this case

the three. So we do a Times by

R. And then we multiply by, in
this case by three again 3 times

by 6 gives 18, so we multiply by
R again, AR squared.

And then we multiply by three
again to give us the 54.

So by our again AR cubed.

And what's our end term in this
case? While A is the first term

8 times by R, is the second term
8 times by R-squared is the

third term 8 times by R cubed?
Is the fourth term, so it's a

times by R to the N minus one.
Because this power there's a

one. There is always one less.

And the number of the term,
then its position in the

sequence. And this is the end
term, so it's a Times my R to

the N minus one.

What about adding up a
geometric progression? Let's

write that down. SN is
equal to a plus R

Plus R-squared Plus.

Plus AR to the N minus one,
and that's the sum of N terms.

Going to use another trick
similar but not the same to what

we did with arithmetic
progressions. What I'm going to

do is I'm going to multiply
everything by the common ratio.

So I've multiplied SN by are
going to multiply this one by R,

but I'm not going to write the
answer there. I'm going to write

it here so I've a Times by R and
I've written it there plus now I

multiply this one by R and that
would give me a R-squared. I'm

going to write it there.

So that term is being multiplied
by R and it's gone to their

that's being multiplied by R and
it's gone to their. This one

will be multiplied by R and it
will be a R cubed and it will

have gone to their.

Plus etc plus, and we think
about what's happening.

That term will come to here and
it will look just like that one.

Plus and then we need to
multiply this by R, and that's

another. Are that we're
multiplying by, so that means

that becomes AR to the N.

Now look at why I've lined these
up AR, AR, AR squared. Our

squared, al, cubed, cubed and so

on. So let's take these two
lines of algebra away from each

other, so I'll have SN minus R
times by SN is equal to. Now

have nothing here to take away
from a, so the a stays as it is.

Then I've AR takeaway are, well,
that's nothing. A R-squared

takeaway R-squared? That's
nothing again, same there. And

so on and so on. AR to the N

minus one. Take away a art. The
end minus one nothing and then

at the end I have nothing there

take away.
AR to the N.

Now I need to look closely at
both sides of what I've got

written down, and I'm going to
turn this over and write it down

again. So we've SN minus
RSN is equal to A.

Minus AR to the N.

Now here I've got a common
factor SN the some of the end

terms when I take that out, I've
won their minus R of them there,

so I get SN times by one minus R
is equal 2 and here I've got a

common Factor A and I can take a
out giving me one minus R to the

N. Remember it was the sum of N
terms that I wanted so.

SN is equal to a Times 1 minus R
to the N and to get the SN on

its own, I've had to divide by
one minus R, so I must divide

this by one minus R.

And that's my formula for the
sum of N terms of a geometric

progression. And let's just
remind ourselves what the

symbols are N is equal to the
number of terms.

A is the first
term of our

geometric
progression and are

we said was called
the common ratio.

OK, and let's just remember the
NTH term in the sequence was AR

to the N minus one. So those
are our fax so far about GPS

or geometric progressions. Let's
see if we can use these facts

in order to be able to help
us solve some problems and do

some questions. So first of all,
let's take this 2 + 6 +

18 + 54 plus. Let's say there
are six terms. What's the answer

when it comes to adding those

up? Well, we know that a
is equal to two. We know that

our is equal to three and we
know that N is equal to six. So

to solve that, all we need to do
is write down that the sum of N

terms is a Times 1 minus R to
the N all over 1 minus R.

Substitute our numbers in two

times. 1 - 3
to the power 6.

Over 1 -

3. So this is 2 * 1 -
3 to the power six over minus

two, and we can cancel a minus
two with the two that we leave

as with a minus one there and
one there if I multiply

throughout by the minus one,
I'll have minus 1 * 1 is minus

one and minus one times minus 3
to the six is 3 to the 6th, so

the sum of N terms is 3 to the

power 6. Minus one and with a
Calculator we could workout what

3 to the power 6 - 1

was. Let's take

another. Question to do with
summing the terms of a geometric

progression. What's the sum
of that? Let's say for five

terms. While we can begin by
identifying the first term,

that's eight, and what's the
common ratio?

Well, to go from 8 to 4 as a
number we would have it, but

there's a minus sign in there.
So that suggests that the common

ratio is minus 1/2. Let's just
check it minus four times. By

minus 1/2 is plus 2 + 2 times Y
minus 1/2 is minus one, and we

said five terms, so Ann is equal

to 5. So we can write
down our formula. SN is equal to

a Times 1 minus R to the
power N all over 1 minus R.

And so A is 8.

1 minus and this is
minus 1/2 to the power

5. All over 1 minus minus
1/2. You can see these

questions get quite
complicated with the

arithmetic, so you have to
be very careful and you

have to have a good
knowledge of fractions.

This is 8 * 1. Now let's have
a look at minus 1/2 to the power

5. Well, I'm multiplying the
minus sign by itself five times,

which would give me a negative
number, and I've got a minus

sign there outside the bracket.
That's going to mean I've got 6

minus signs together. Makes it
plus. So now I can look at the

half to the power 5.

Well, that's going to be one

over. 248-1632 to
to the power

five is 32.

All over 1 minus minus 1/2.
That's 1 + 1/2. Let's write

that as three over 2.

So this is equal to.

Now I've got 8.

Times by one plus, one over 32,
and I'm dividing by three over 2

to divide by a fraction. We
invert the fraction that's two

over 3 and we multiply by and we
just turn the page to finish

this one off.

So we have SN is
equal to 8 * 1

+ 1 over 32 times
by 2/3 is equal to

8 times by now one
and 132nd. Well, there are

3230 seconds in one, so
altogether there I've got 33.

30 seconds times

by 2/3. And we
can do some canceling threes

into 30. Three will go 11 and
threes into three. There goes

one. Twos into two goes one and
tools into 32, goes 16 and 18

two eight goes one and eight
into 16 goes 2. So we 1 *

11 * 1 that's just 11 over 2
because we've 2 times by one

there. So we love Nova two or we
prefer five and a half.

So that we've got the some of
those five terms of that

particular GP. Five and a half,
11 over 2 or 5.5.

But here's a different question.
What if we've got the sequence

248? 128 how many terms are
we got? How many bits do we

need to get from 2 up to
128? Well, let's begin by

identifying the first term

that's two. This is.

A geometric progression because
we multiply by two to get each

term. So the common ratio are is
2 and what we don't know is

what's N. So let's have a look.
This is the last term and we

know our expression for the last
term. 128 is equal to AR to the

N minus one.

So let's substituting some
of our information.

A is 2 times by two
4R to the N minus one.

Well, we can divide both
sides by this two here,

which will give us.

64 is equal to two to the
N minus one.

I think about that it's 248
sixteen 3264 so I had to

multiply 2 by itself six times
in order to get 64, so 2

to the power 6, which is 64
is equal to 2 to the power

N minus one, so six is equal
to N minus one, and so N

is equal to 7, adding one.

To each side. In other words,
there were Seven terms in our.

Geometric progression. Type
of question that's often given

for geometric progressions is
given a geometric progression.

How many terms do you need
to add together before you

exceed a certain limit? So, for
instance, here's a geometric.

Progression. How many times of
this geometric progression do we

need to act together in order to
be sure that the some of them

will get over 20?

Well, first of all, let's try
and identify this as a geometric

progression. The first term is
on and it looks like what's

doing the multiplying. The
common ratio is 1.1. Let's just

check that here.

1.1 times by one point, one
well. That's kind of like 11 *

11 is 121.

With two numbers after the
decimal point in one point 1 *

1.1 and with two numbers after
the decimal point there. So yes,

this is a geometric progression.

So let's write down our formula
for N terms sum of N terms

is equal to a Times 1 minus
R to the N.

All over 1 minus R. We want to
know what value of N is just

going to take us over 20.

So let's substituting some
numbers. This is one for

a 1 - 1.1 to

the N. All over
1 - 1.1 that

has to be greater
than 20.

So one times by that isn't going
to affect what's in the

brackets. That would be 1 - 1.1
to the N all over 1 - 1.1

is minus nought. .1 that has to
be greater than 20.

Now if I use the minus sign
wisely. In other words, If I

divide if you like.

Minus note .1 into there as

a. Division, then I'll have.

The minus sign will make that a
minus and make that a plus, so

I'll have one point 1 to the N
minus one and divided by North

Point one is exactly the same as
multiplying by 10. That means

I've got a 10 here.

That I can divide both sides by.

So let's just write this down
again 1.1 to the N minus one

times by 10 has to be greater
than 20. So let's divide both

sides by 10, one point 1 to the
N minus one has to be greater

than two and will add the one to
both sides 1.1 to the end has to

be greater than three.

Problem how do we find N? One of
the ways of solving equations

like this is to take logarithms
of both sides, so I'm going to

take natural logarithms of both
sides. I'm going to do it to

this site first. That's the
natural logarithm of 3 N about

this side. When you're taking a
log of a number that's raised to

the power, that's the equivalent
of multiplying the log of that

number. By the power that's N
times the log of 1.1. Well

now this is just an equation
for N because N has got to

be greater than the log of 3
divided by the log of 1.1

because after all.

Log of three is just a number
and log of 1.1 is just a number

and this is the sort of
calculation that really does

have to be done on a Calculator.

So if we take our Calculator and
we turn it on.

And we do the calculation. The
natural log of three.

Divided by the natural log of
1.1, we ask our Calculator to

calculate that for us. It tells
us that it's 11.5 to 6 and some

more decimal places. We're not
really worried about these

decimal places. An is a whole
number and it has to be greater

than 11 and some bits, so N has
got to be 12 or more.

That's one last twist
to our geometric progression.

Let's have a look at

this one. What have
we got got

a geometric progression.
First term a

is one.

Common ratio is 1/2 because
we're multiplying by 1/2 each

time. That write down

some sums. S1, the sum
of the first term is just.

1. What's

S2? That's the
sum of the first 2 terms, so

that's. Three over
2.

What's the sum of the first
three terms? That's one.

Plus 1/2 +

1/4. Add those up
in terms of how many quarters

have we got then that is

7. Quarters As
for the sum of

the first.

4. Terms.
Add those up in terms of how

many eighths if we got so we've
got eight of them there. Four of

them there. That's 12. Two of
them there. That's 14 and one of

them there. That's 15 eighths.

Seems to be some sort of
pattern here.

Here we seem to be 1/2 short of

two. Here we
seem to

be 1/4. Short of two
here, we seem to be an eighth

short of two and we look at the
first one. Then we're clearly 1

short of two.

He's a powers of two. Let's have
a look 2 - 2 to the

power zero, 'cause 2 to the
power zero is 1 two.

Minus.

2 to the power minus one 2
- 2 to the power minus two

2 - 2 to the power minus

three. But each of these is
getting smaller. We're getting

nearer and nearer to two. The
next one we take away will be a

16th, the one after that will be
a 32nd and the next bit we take

off 2 is going to be a 64th and
then a 128 and then at one

256th. So we're getting the bits
were taking away from two are

getting smaller and smaller and
smaller until eventually we

wouldn't be able to distinguish
them from zero.

And so if we could Add all of
these up forever, a sum to

Infinity, if you like the
answer, or to be 2 or as near as

we want to be to two. So let's
see if we can have a look at

that with some algebra.

We know that the sum to end
terms is equal to a Times 1

minus R to the N all over
1 minus R.

What we want to have a look at
is this thing are because what

was crucial about this?

Geometric progression was at the
common ratio was a half a

number less than one.

So let's have a look what

happens. When all is bigger than
one to R to the power N.

We are is bigger than one and we
keep multiplying it by itself.

Grows, it grows very rapidly and
really gets very big very

quickly. Check it with two, 2,
four, 816. It goes off til

Infinity. And because it goes
off to Infinity, it takes the

sum with it as well.

What about if our is equal

to 1? Well, we can't really
use this formula then because we

would be dividing by zero. But
if you think about it, are

equals 1 means every term is the
same. So if we start off with

one every term is the same 1111
and you just add them all up.

But again that means the sum is
going to go off to Infinity if

you take the number any number
and add it to itself.

An infinite number of times
you're going to get a very, very

big number. What happens if
our is less than minus one?

Something like minus 2?

Well, what's going to happen
then to R to the N?

Well, it's going to be plus an.
It's going to be minus as we

multiply by this number such as
minus two. So we have minus 2 +

4 minus A. The thing to notice
is it's getting bigger, it's

getting bigger each time. So
again are to the end is going to

go off to Infinity. It's going
to oscillate between plus

Infinity and minus Infinity, but
it's going to get very big and

that means this sum is also
going to get.

Very big.

What about our equals minus one?
Well, if R equals minus one,

let's think about a sequence
like that. Well, a typical

sequence might be 1 - 1.

1 - 1 and we can see the
problem. It depends where we

stop. If I stop here the sum is
0 but if I put another one

there, the sum is one. So we've
got an infinite number of terms

then. Well, it depends on money
I've got us to what the answer

is so there isn't a limit for
SN. There isn't a thing that it

can come to a definite number.

Let's have a look. We've
considered all possible values

of our except those where are is
between plus and minus one.

Let's take our equals 1/2
as an example.

Or half trans by half is 1/4.

Reply by 1/2. Again
that's an eighth.

Multiply by 1/2 again, that's a

16. Multiplied by 1/2 again,
that's a 32nd.

By half again that's a 64th by
1/2 again, that's 128.

It's getting smaller, and if we
do it enough times then it's

going to head off till 0.

What about a negative one? You
might say, let's think about

minus 1/2. Now multiplied by
minus 1/2, it's a quarter.

Multiply the quarter by minus
1/2. It's minus an eighth.

Multiply again by minus 1/2.
Well, that's plus a 16th.

Multiply again by minus 1/2.
That's minus a 32nd, so we're

approaching 0, but where dotting
about either side of 0 plus them

were minus, then were plus then
where mine is.

We're getting nearer to zero
each time, so again are to the

power. N is going off to zero.
What does that mean? It means

that this some. Here we can have
what we call a sum to Infinity.

Sometimes it's just written with
an S and sometimes it's got a

little Infinity sign on it.

What that tells us? Because this
art of the end is going off to

0 then it's a times by one over
1 minus R and that's our sum to

Infinity. In other words, we can

add up. An infinite number of
terms for a geometric

progression provided. The common
ratio is between one and minus

one, so let's have a look
at an example. Supposing we've

got this row.

Metric progression.
Well, first term is one

now a common ratio is
1/3.

And what does this come to when

we add up? As many terms as
we can, what's the sum to

Infinity? We know the formula
that's a over 1 minus R, so

let's put the numbers in this
one for a over 1 - 1/3.

So the one on tops OK and the
one minus third. Well that's

2/3, and if we're dividing by a
fraction then we invert it and

multiply. So altogether that
would come to three over 2, so

it's very easy formula to use.

Finally, just let's recap for a

geometric progression. A.

Is the first term.

Aw.

Is the common.

Ratio. So a
geometric progression looks like

AARA, R-squared, AR, cubed and
the N Terminus series AR

to the N minus one.

And if we want to add up this
sequence of numbers SN.

Then that's a Times 1 minus R to
the power N or over 1 minus R.

And if we're lucky enough to
have our between plus and

minus one, sometimes that's
written as the modulus of art

is less than one. If we're
lucky to have this condition,

then we can get a sum to
Infinity, which is a over 1

minus R.