### Video Transcript

We’re going to take a look at some
arithmetic sequences. We’ll take a look at the
definition. And we’ll see some examples and go
through some typical arithmetic sequence questions. We’ll be analysing the differences
between consecutive terms in the sequence and looking for the value of the zeroth
term in order to find the formula for the 𝑛th term of the sequence.

An arithmetic sequence then is a
sequence of numbers in which you add the same amount to each term to get the next
term, for example, five, eight, 11, 14, and 17. The first term of this sequence is
five, the second term is eight, the third term is 11, the fourth term is 14, the
fifth term is 17, and so on. So my little orange numbers in
circles are just telling me the position of the number within that sequence. And the big blue numbers are the
values of those terms. And you can see that as we move
through that sequence, every time we move to the next sequence, it’s three bigger
than in the previous term in the sequence.

So in this case, we’d say that
consecutive terms in the sequence have a common difference of positive three. So it’s that having a common
difference of something which makes it in an arithmetic sequence. Then we have a first term. And in that case, our first term
has a value of five. Knowing the value of the first term
and the common difference means that you can work out the value of any term in that
sequence. So when you spot that pattern, that
consecutive terms have a common difference, then you can say it’s an arithmetic
sequence. So when you know what an arithmetic
sequence is, you can answer questions like this.

Write the next three terms in the
arithmetic sequence 12, 19, 26, 33.

So we’ve been told the first,
second, third, and fourth terms in the sequence. And we wanna work out what the
fifth, sixth, and seventh terms are. Well, to get from 12 to 19, we
needed to add seven. To get from 19 to 26, we also
needed to add seven. And to get from 26 to 33, yet again
we needed to add seven. So the common difference is
seven. So to get from the fourth to the
fifth term, we just need to add seven. And 33 plus seven is 40. Add seven again to get the sixth
term. 40 plus seven is 47. And then add seven again to get the
seventh term, 54. So our answer is that the next
three terms are 40, 47, and 54.

Let’s try another one then.

Write the next three terms in the
sequence 3.6, 4.3, 5.0, 5.7, and so on.

This means that our first term is
3.6, our second term is 4.3, the third term is 5.0, and the fourth term is 5.7. We want the next three terms. So that’s the fifth, sixth, and
seventh terms. Now when I look at consecutive
terms to get from 3.6 to 4.3, I just add 0.7. To get from 4.3 to 5.0, I add
0.7. And to get from 5.0 to 5.7, I just
add 0.7. So they’ve got a common difference
of 0.7.

So this means we’ve got our
arithmetic sequence. So I can just add 0.7 again
repeatedly in order to get the fifth, sixth, and seventh terms. And that will give me 6.4, 7.1, and
7.8. So that’s our answer.

And we can see that arithmetic
sequences can involve integer numbers. They can involve decimal numbers,
fractions. So long as the difference between
consecutive terms is always the same, it’s an arithmetic sequence. Okay, let’s try this question
then.

Write the 10th term in the
arithmetic sequence 23, 19, 15, 11, seven, and so on.

Well, now we’ve been given five
terms in the sequence. And we’re not worried about the
sixth, seventh, eighth, and ninth, but we want to know what the 10th term is. Now first, let’s look at the common
difference. To go from 23 to 19, we have to
take away four. From 19 to 15, take away four. 15 to 11, take away four. 11 to seven, take away four. So our common difference is
negative four. So it’s even okay for our common
differences to be negative.

Now if we think about it, there are
two different ways to work out what the 10th term is. We could either work out the sixth,
seventh, eighth, and ninth and then work out the 10th. Or we could say that, from seven,
we’ve got to take away five lots of negative four for the five steps that take us up
to the 10th term.

So doing that the first way, seven
take away four is three. Three take away four is negative
one. Negative one take away another four
is negative five. And negative five take away another
four is negative nine. And negative nine take away another
four is negative 13. Or as I say, we could’ve started at
seven and taking away five lots of four so that’s seven take away 20, which is
negative 13. So either way, we’d have got the
same answer for our 10th term.

But this is beginning to become a
bit cumbersome. What if the question asked us to
write the 250th term of the sequence? We wouldn’t wanna work out all the
different terms up to the 250th term. So let’s look at a way of coming up
with a general formula for the 𝑛th term of the sequence and then using that to
predict what the 𝑛th term is going to be.

So let’s type the sequence three,
seven, 11, 15, and so on. Let’s label our terms one, two,
three, four. And our common difference is
positive four. We have to add four to each term to
get the next. Now I’m trying to come up with a
formula for the general term in this sequence. So I’m gonna put some letters in
here. Let’s let the position in our
sequence be 𝑛. So if 𝑛 equals one, we’re talking
about the first number in the sequence. If 𝑛 is two, we’re talking about
the second number, and so on. And I’m gonna use this notation,
𝑡, and then in parentheses 𝑛 to represent the 𝑛th term. So when 𝑛 is one, 𝑡 one is
three. When 𝑛 is two, 𝑡 two is equal to
seven; 𝑡 three is equal to 11; 𝑡 four, the fourth term, is equal to 15; and so
on.

Now if we look at this carefully,
every time my position in the sequence goes up by one, the value of the term goes up
by four. So the value of the term is going
up four times faster than the position. And since 𝑛 is the value of our
position, our formula must have positive four times 𝑛 in it somewhere. So let’s work out that 𝑛 is one
for the first position, 𝑛 is two for the second, 𝑛 is three for the third, 𝑛 is
four for the fourth, and so on. Let’s work out. If we said our formula was just
positive four times 𝑛, what values would we generate in our sequence?

Well, positive four times — well,
𝑛 is one in this case for our first term — positive four times one will just be
four. When 𝑛 is two, we’re talking about
the second term in the sequence. So positive four times two would be
eight. For the third term in the sequence,
we’d be doing positive four times three, which will be 12. And for the fourth term, we’d be
doing positive four times four, which will be 16. So if our formula was positive four
times 𝑛, the sequence we would generate would be four, eight, 12, 16. But that’s not the values we’re
looking for.

We got four. We wanted three. We got eight. We wanted seven. We got 12. We wanted 11. We got 16. We wanted 15. Well, what have I got to do to each
of these numbers in order to turn the number- turn the numbers that I want?

Well, four take away one gives me
three. Eight take away one gives me
seven. 12 take away one would give me
11. And 16 take away one would give me
15. So if I take these four 𝑛-values
and take away one, this process generates exactly the numbers I’m looking for. So the formula for the 𝑛th term of
the sequence is that 𝑛th term 𝑡𝑛 is equal to positive four 𝑛. Well, we don’t normally bother
writing a positive if it’s positive, so four 𝑛 minus one.

And the 250th term of this
sequence, 𝑡 250, will just be four times 250 minus one. Well, four times 250 is 1000, take
away one is 999. So the 250th term will be 999. We didn’t have to work out the
fifth, the sixth, the seventh, and so on in order to work out the 250th term.

Even better, we can go on to work
out the 1527486th term in the sequence just by plugging in that value for 𝑛 in our
formula. And when we do that, we get an
answer of 6109943. Now that’s saved us a lot of time
working and not having to work out millions of terms in order to work out the term
that we’re interested in.

Now it’s also worth pointing out
there’re different ways of expressing this. Some people say 𝑡𝑛, so 𝑡 with a
little 𝑛 underneath it, or 𝑎𝑛 for the 𝑛th term of the sequence. And some people use 𝑡 and then in
the parentheses 𝑛. Okay, let’s try out this method
then.

Find the 952nd term in the sequence
107, 99, 91, 83, and so on.

So we’ll write out our 𝑛-values,
our positions in the sequence, one, two, three, four, and so on, and our 𝑛th
terms. So the first term is 107; the
second term, 𝑡 two, is 99; the third term, 𝑡 three, is 91; and so on. And in moving from term to term,
we’re taking away eight each time. So let’s try negative eight 𝑛 for
our formula.

And to do that, we’re just gonna
substitute in the values of 𝑛 one, two, three, and four into that formula. So it’s gonna be negative eight
times one, negative eight times two, negative eight times three, negative eight
times four, and so on. So that gives us negative eight,
negative 16, negative 24, and negative 32. Now they’re clearly not the numbers
that we’re looking for. We got negative eight, but we
wanted 107. We got negative 16, but we wanted
99. We got negative 24, but we wanted
91. And we got negative 32, but we
wanted 83. So we clearly got quite the wrong
numbers.

But if I add 115 to each of the
numbers we got, we do get the numbers that we wanted. So our formula is negative eight 𝑛
plus 115. So the formula for our 𝑛th term,
𝑡𝑛, is negative eight 𝑛 plus 115. And when we substitute in the
number 952 to find the 952nd term, the formula gives us negative eight times 952
plus 115, which is negative 7501.

Now all of this is great, but it
does mean that we gotta do a lot of working out and write a lot of things down. So we’re gonna look for a slightly
shorter way of doing this. So if we plotted our 𝑛-values as
𝑥-coordinates and the corresponding terms themselves as the 𝑦-coordinates, then we
get a table of values that looks something like this.

Remember, the 𝑥s are really the
𝑛-values we’ve just been looking at, and the 𝑦s are the 𝑡𝑛s, the 𝑛th terms. And if we plotted those points,
they’d look something like that. Now 𝑛 can only take integer values
one, two, three, four, and so on because that sequence has only got terms one, two,
three, and four, and so on. It doesn’t really make much sense
to talk about the 3.75th term for the sequence, for example. But if we did join up those points
and then just sort of extend them back to zero, they would cross here at 115. And that’s because every time I
increase my 𝑥-coordinate by one, my 𝑦-coordinate decreases by eight. So if I go from this point here and
decrease my 𝑥-coordinate by one, my 𝑦-coordinate is gonna go up by eight.

So now we have all the information
we need to work out the equation of that line. The slope is negative eight or the
gradient, and the 𝑦-intercept is 115. So the equation is 𝑦 equals
negative eight 𝑥 plus 115. Now we can use this as a shortcut
for working out the formula of our 𝑛th term. So the common difference told us
the slope of that line. And the 𝑦-intercept tells us the
number that we have to add on to the negative eight 𝑥 to get the 𝑛th term in the
formula. Okay, let’s do another question and
see that in action and see how much quicker it makes that question to do.

Find the formula for the 𝑛th term
of the arithmetic sequence 6.8, 7.9, 9.0, 10.1, and so on.

So we’ve still got our 𝑛s: one,
two, three, four, the positions in the sequence. And we’ve still got our terms, our
𝑡𝑛s, 6.8, 7.9, 9.0, 10.1, and so on. Now looking at the common
difference from term to term, to get to each consecutive term, we need to add
1.1. So that tells us the multiple of
𝑛. So what we need to do is find the
value of our zeroth term, the 𝑦-intercept values. So we’re looking for 𝑛 equals
zero.

Now when we were moving along
consecutively our terms, we were adding 1.1 to each term. So if we’re going back from the
first term to the zeroth term, we’ve got to subtract 1.1. And 6.8 minus 1.1 is 5.7. So the formula for our 𝑛th term is
the common difference positive 1.1 times 𝑛. Since it’s positive, we won’t
bother writing the positive sign. And we then have to add on 5.7. That’s a positive 5.7. So there’s our formula. The 𝑛th term is 1.1𝑛 plus
5.7.

And if the question asked us now to
find the 88th term in the sequence, the calculation we’d have to do to find out 𝑡
88 is 1.1 times 88. So 88 is the value of 𝑛, the 88th
term, plus 5.7, which is 102.5. So the 88th term in our sequence
would be 102.5.

So to summarise, the formula for
the 𝑛th term of an arithmetic sequence is always of this format. The 𝑛th term is equal to something
times 𝑛 plus something. Now the 𝑎, the something we’re
timesing 𝑛 by, is the common difference of each term in the arithmetic
sequence. And the 𝑏-value would be the value
of the zeroth term in the sequence. And an arithmetic sequence, 𝑛th
term formula, will always be in that format. If it’s got squareds or roots or
powers of 𝑛 in it, it can’t be any of those formulae. They’re not linear, so they can’t
represent an arithmetic sequence.

Now one more sort of question that
you might see in relation to arithmetic sequences is like this. Does 117 belong to the arithmetic
sequence five, 18, 31, 44, and so on? And the way you approach these is
much the same in the first instances, the ones we’ve just been looking at.

First, we’re gonna analyse the
common difference. And in this case, that’s plus 13
each time. So that tells us the multiple of 𝑛
in our formula. And then we’re gonna find the
zeroth term of that sequence. So moving back from the first to
the zeroth term, we’re moving backwards. So we’re gonna do the opposite of
adding 13. That’s taking away 13. Our first term was five, so five
take away 13 is negative eight. So our formula for the 𝑛th term
then is the common difference times 𝑛. So that’s gonna be 13𝑛. And the zeroth term is the thing
that we’re gonna be adding or taking away in this case at the end. So 𝑡𝑛 is 13𝑛 minus eight.

Now remember, we were saying before
that the 𝑛-values must all be whole numbers. We only talk about the first term
in the sequence, the second, the third, and so on. We don’t talk about the three and a
half-th term. So if 117 is in the term, then the-
there’s gonna be an 𝑛th term that is 117. And the 𝑛-value corresponding to
that would be an integer, a whole number. So there’s gonna be 117 is equal to
13 times some number minus eight. Well, if I add eight to both sides
of that equation, I get 125 is equal to 13𝑛. And now if I divide both sides by
13, I get 𝑛 is equal to 125 over 13. So that’s nine and
eight-thirteenths. That’s not a whole number. So our answer is 117 is not an
exact term in the sequence. It would be the nine and
eight-thirteenth term. That’s not an integer, so it’s not
a term.

Okay, one last quick question
then. Does 5.2 belong to the arithmetic
sequence with the general term formula 𝑡𝑛 is equal to 106 minus 1.8𝑛?

And this again boils down to the
fact that if the value of 𝑛 that generates a term of 5.2 is an integer, then yes it
does belong to the sequence. But if the value of 𝑛 that
generates that term isn’t an integer, then it doesn’t. So we’re putting 𝑡𝑛 equal to
5.2. Now in this case, it’s negative
1.8𝑛 there. So I’m gonna add 1.8𝑛 to both
sides to give me a positive number of 𝑛s. Then I can subtract 5.2 from each
side. And finally divide by 1.8 to leave
me with one 𝑛 on the left-hand side. And in this case, that gives me a
whole number answer, 56. So our answer is, yes, 5.2 is the
56th term in the sequence.