### Video Transcript

In this video, weβll discover how
to represent a sequence as a function of a positive integer variable called an
index. Weβll look at a mixture of
questions on both arithmetic and geometric sequences, including how to find the πth
term of these sequence types and also how to write a sequence given the πth
term. But before we begin some questions,
letβs recap some of the terminology around sequences and types of sequence we might
encounter.

Firstly, we have that a sequence is
an ordered list of terms. Terms are usually named either π
sub π or π sub π, where π or π are the index. In this video, weβll tend to use π
as the index. So, for example, π sub three would
be the third term. Sometimes sequences start with the
index π as π equals one, and sometimes they start with π equals zero. In this case, the sequence will
begin with what we would call the zeroth term. And then weβd have the first term
and the second term and so on. We usually get a bit of a clue in
question types if they ask us to give the πth term where π is greater than or
equal to zero. In this case, we know that the
index should start with π equals zero. The terms in a sequence can be
given as a list or defined by a rule often related to the index.

Letβs now recap how we would find
the πth term of an arithmetic sequence and of a geometric sequence. An arithmetic sequence is a
sequence with the difference between two consecutive terms constant. This difference is called the
common difference. An example of an arithmetic
sequence would be the sequence five, 10, 15, 20, and so on, where we can see that
the common difference between any two consecutive terms would be positive five.

In order to find the πth term or
π sub π of an arithmetic sequence, we calculate π, the first term, plus π minus
one times d, where d is the common difference. A geometric sequence is a sequence
with the ratio between two consecutive terms constant. And this ratio is called the common
ratio. An example of a geometric sequence
might be the sequence two, four, eight, 16, and so on. We can find the common ratio of a
geometric sequence by taking any term and dividing it by the term before it. For example, if we divided the
fourth term of 16 by the third term of eight, weβd get the value of two for the
ratio. Weβd also get this ratio of two if
we divided eight by four.

To find the πth term of a
geometric sequence π sub π, we calculate π multiplied by π to the power of π
minus one, where π is the first term and π is the common ratio. Now that weβve recapped what we
know about sequences, letβs have a look at the first question, where weβll need to
establish which type of sequence we have and then find its πth term.

In any sequence pattern, if the
difference between any two successive terms is a fixed number π, then this is
an arithmetic sequence. Consider the sequence one,
four, seven, 10, and so on and then answer the following questions. Is the sequence arithmetic? What is the value of π? What is the general term of
this sequence with π is greater than or equal to zero?

To start this question, weβre
given a little reminder about the definition of an arithmetic sequence. And itβs one which has a
difference or common difference between any two successive terms the same. This word successive is like
the word consecutive. It simply means two terms where
one immediately follows the other one. We are asked to consider the
sequence one, four, seven, 10, and so on. If we look at the first
question, weβre asked if the sequence is arithmetic. So if itβs arithmetic, it will
have a common difference between any two consecutive terms.

So if we wanted to find the
difference between the first and the second term, we would work out four take
away one, which of course is three. To find the next common
difference between the third term and the second term, we would work out seven
take away four, and thatβs also three. In the same way, the difference
between 10 and seven is also three. As we have a common difference,
then we do have an arithmetic sequence. And we can say yes as the
answer for the first part of this question. The second part of the question
asks us, what is the value of π? π is the common difference, and weβre
reminded of that in the question text. And itβs also nice and easy to
work out; itβs three. And thatβs the second part of
the question answered.

In the final part of this
question, weβre asked to find the general term of this sequence with π greater
than or equal to zero. Remember that the general term
is often seen as the πth term. The fact that π is greater
than or equal to zero means that the index of this sequence should start with π
equals zero. So actually, this sequence
begins with a zeroth term. Then weβd have the first term,
then the second term and the third term and so on. We should recall that there is
a formula to help us find the πth term of an arithmetic sequence. The πth term π sub π is
equal to π plus π minus one multiplied by π, where π is the first term and
π is the common difference.

When we look at the sequence
one, four, seven, 10, and so on, the first term is really the same as π sub
one. So the value of π, which we
plug in, will be four. And as the value of π, which
we worked out earlier, is three, then we add on π minus one multiplied by
three. When we distribute the three
across the parentheses, we get three multiplied by π, which is three π, and
three multiplied by negative one, which is negative three. Finally, when we simplify this,
we get four minus three, which is one, plus three π or three π plus one. And thatβs the answer for the
third part of this question: the general term or πth term of this sequence is
three π plus one.

But it is important to note
that this was because the index π was greater than or equal to zero. If the index had started with
one, i.e., π is greater than or equal to one, then we would have worked out the
πth term to be three π minus two. So itβs really important to
read the question to see if thereβs an indication that the index, in fact,
begins with zero. In this case, however, we can
give the general term as three π plus one.

Weβll now take a look at another
question.

In a geometric sequence, the
ratio between any two successive terms is a fixed ratio π. Consider the sequence one-half,
one-quarter, one-eighth, one sixteenth, and so on. Is this sequence geometric? Consider the sequence one-half,
one-quarter, one-eighth, one sixteenth, and so on. What is the value of π? Consider the sequence one-half,
one-quarter, one-eighth, one sixteenth, and so on. What is the general term of
this sequence?

In this question, weβre given a
reminder of what a geometric sequence is. Itβs one which has a fixed
ratio or common ratio between any two successive terms. In each of the three parts of
this question, weβre considering the same sequence. And in the first part of this
question, weβre asked if this given sequence is geometric. So letβs write down this
sequence. If it is geometric, then there
will be a common ratio π between any two consecutive or successive terms. So letβs see if we can find a
ratio between the first two terms, one-half and one-quarter. To find the ratio, we take the
second term of one-quarter and divide it by one-half.

When weβre dividing fractions,
we write the first fraction and we multiply it by the reciprocal of the second
fraction. We can take out a common factor
of two from the numerator and denominator. And then multiplying the
numerators gives us one, and multiplying the denominators gives us two. That means that the ratio
between the first and second term is one-half. Next, we will find the ratio
between the second and third term of one-quarter and one-eighth, so weβll
calculate one-eighth divided by one-quarter. Multiplying by the reciprocal
of the fraction one-quarter, we calculate one-eighth multiplied by four over
one. And so, once again, we get the
ratio of one-half.

It looks like we probably do
have a common ratio, but itβs always worth checking all the terms to make sure
that there is a common ratio on all of them. And when we calculate one
sixteenth divided by one-eighth, we also get the ratio of one-half. Therefore, we have established
that this sequence does have a fixed or common ratio, and so it must be
geometric. And we can say yes as the
answer for the first part of this question. In the second part of this
question, weβre asked to find the value of π for the same sequence. Remember that π is the fixed
ratio and itβs nice and simple. We have just worked it out to
be one-half, which is the second part of this question answered.

In the final part of this
question, weβre asked to find the general term of this sequence. And we can remember that the
general term is another way of asking for the πth term. If we started with the first
term written as π sub one equal to one-half, then the second term would be π
sub two and it would be one-quarter. The third and fourth terms can
be written as π sub three and π sub four. So when weβre finding the
general term, weβre really looking for the rule that would allow us to work out
the πth term or π sub π.

We can remember that there is a
general formula to allow us to work out the πth term of any geometric
sequence. π sub π is equal to π
multiplied by π to the power of π minus one, where π is the first term and π
is the common ratio. The values that we need to plug
into the formula will be π equals one-half, as that was the first term, and π
we worked out as one-half. Therefore, the πth term of
this sequence can be given as a half multiplied by one-half to the power of π
minus one. When we give our answer, itβs
important that we indicate the values of π. When we find the πth term or
general term, we started with π equals one. So the answer for the third
part of the question is that the general term of the sequence is one-half
multiplied by one-half to the power of π minus one for values of π greater
than or equal to one.

We can, of course, further
simplify this general term using one of the rules of exponents. If we consider that the first
value of one-half is equivalent to one-half to the power of one, and so adding
the exponents one and π minus one will give us simply the exponent of π. If we then plugged in the
values of π equals one, two, three, or four into either of these formulas, weβd
get the first four terms of the sequence that we were given, and so verifying
that we have the correct answer for the general term.

Next, letβs have a look at a
question where we need to find the first three terms of the sequence given the
general term.

Find the first three terms of
the sequence whose general term π sub π equals π over π plus one.

In this question, weβre given a
general term or an πth term for a sequence as π over π plus one. This value of π is an index
for the sequence. So when we want to work out the
first term, weβd actually be working out the value of π sub one, which means
that we plug in the value of one for every value of π. Therefore, weβll have π sub
one is equal to one over one plus one. Simplifying this gives us the
value of one-half.

To find the second term or the
value of π sub two, weβll be plugging in the value of π equals two. This gives us two over two plus
one, which when we simplify it gives us the value of two-thirds for the second
term. Finally, for the third term,
weβll be working out the value of π sub three. So we plug in π equals three
into the general term. And we have three over three
plus one, which simplifies to three-quarters. Therefore, we can give the
answer that the first three terms of this sequence must be one-half, two-thirds,
and three-quarters.

Weβll now have a look at one final
question.

Consider the sequence one, one,
three-quarters, four-eighths, and so on. Which of the following is the
general term of this sequence such that π is greater than or equal to zero? Option (A) π over two to the
power of π. Option (B) π minus one over
two to the power of π. Option (C) π plus one over two
to the power of π. Option (D) two π over two to
the power of π. Or option (E) π plus two over
two to the power of π.

In this question, weβre given a
sequence and asked to find its general term. When weβre finding a general
term, weβre really finding a rule that connects the term number with the actual
value of the term. When weβre given that the index
π is greater than or equal to zero, that means that our sequence begins with
the zeroth term. We then have the first term π
sub one, the second term π sub two, and so on. The πth term would be π sub
π. So given any value of π, what
would the value in the sequence be? If we consider this sequence,
there isnβt a common difference between any two consecutive terms, so this isnβt
an arithmetic sequence. There also isnβt a common ratio
between any two consecutive terms, so this isnβt a geometric sequence
either.

In order to find the general
term of the sequence, weβll have to apply some logic. Letβs take a closer look at
this term, π sub one with the value of one. What if instead of being this
value of one, this value of π sub one was actually a fraction which simplified
to one? In order for a fraction to
simplify to one, the numerator and denominator would have to have the same
value. Letβs say this fraction was
actually two over something, and to simplify to one, it would need to be two
over two. If we think of the zeroth term
π sub zero as instead of just being one as being a fraction of one over one,
now we can see that the numerators actually have quite a nice pattern. They go from one to two to
three to four. The denominators also have a
different pattern. They go from one to two to four
to eight.

Letβs consider the general term
of the numerators and denominators separately for each value of π starting with
π equals zero. Remember that we picked zero
because this was given to us in the question. So for any index π, what will
the numerator be? Well, every value in the
numerator is one more than its index. So the πth term of the
numerator will be π plus one. For the denominators, these
have a pattern which appears to be doubling. In fact, each denominator is a
power of two. It would be two to the power of
π. For example, for the zeroth
term, two to the power of zero gives us one. For the first term, two to the
power of one gives us two and so on. We can now put together the
general term for the numerator and denominator. So the general term of this
sequence is π plus one over two to the power of π, which was the value given
to us in option (C).

Weβll now summarize the key points
of this video. Firstly, we saw that sequences
consist of terms. Then we saw that terms can be
written as π sub π, where π is the index. We also saw how sometimes the index
π begins with zero and sometimes it begins with one. We recapped arithmetic and
geometric sequences and tried to find their πth terms.