### Video Transcript

In this video, weβre going to look
at arithmetic sequences. First, weβll look at the definition
and then consider some typical arithmetic sequences. Weβll be analyzing the differences
between consecutive terms, looking for the value of the zeroth term in order to find
an πth term formula. The πth term formula will allow us
to find any term in the sequence.

So letβs look at the numbers five,
eight, 11, 14, 17. We know this is a sequence as it is
an ordered list of numbers. A term of a sequence is each number
in a sequence. And we number the terms
consecutively. The first term of this sequence is
five. The second term is eight. The third term is 11. And it could continue in this
pattern. This numbering continues
indefinitely as the sequence can continue indefinitely. The term number tells us the
position of that value in the sequence as well. This means we could say 11 is the
third term in the sequence.

As we move from one term to the
next, weβre adding three to the previous term to get the next consecutive term. Weβre adding three every time. This property, adding three each
time, means that this sequence is an arithmetic sequence. In an arithmetic sequence, the
difference between any two consecutive terms is always the same. We can label our sequence as
such. The difference between any two
consecutive terms in an arithmetic sequence is called the common difference. In our example, the common
difference is positive three.

We should note here that the first
term in an arithmetic sequence is important. If we know the first term and the
common difference β in our case, positive three β we can work out any term in the
sequence. The first term and the common
difference are two key bits of information when dealing with any arithmetic
sequence. To sum this up, if you spot a
sequence where consecutive terms have a common difference, you can say itβs an
arithmetic sequence.

Once we know what an arithmetic
sequence is, we can solve questions like this one.

Write the next three terms in the
sequence: 31, 57, 83, 109, continuing.

Weβve been given the first, second,
third, and fourth terms in a sequence, which means we need to find the fifth, sixth,
and seventh term. So we ask ourselves, how do we get
from 31 to 57? We add 26. And 57 plus 26 equals 83. 83 plus 26 equals 109. This makes positive 26 the common
difference. And it means to get from the fourth
to fifth term, weβll need to add 26 to 109. 109 plus 26 equals 135. 135 plus 26 equals 161. And to find the seventh term, we
need to add 26 to the sixth term. 161 plus 26 equals 187. And so we can say that the next
three terms in this sequence are 135, 161, and 187.

Letβs try another one.

Write the next three terms in the
sequence: 3.6, 4.3, 5.0, 5.7.

Weβve been given the first four
terms. The next three terms would be the
fifth, sixth, and seventh term. To go from 3.6 to 4.3, itβs an
increase of 0.7, an increase of seven-tenths. From 4.3 to five is an increase of
0.7, and from five to 5.7 is an increase of 0.7, which means we have a common
difference of positive 0.7 and confirms that this is an arithmetic sequence. So we know to find the fifth term,
we need to add the common difference to the fourth term. 5.7 plus 0.7 equals 6.4, 6.4 plus
0.7 equals 7.1, and 7.1 plus 0.7 equals 7.8. The next three terms in this
sequence would be 6.4, 7.1 and 7.8.

What weβre seeing now is that
arithmetic sequences can occur in all kinds of types of values, in integers and
decimals but also in fractions. As long as the common difference
between consecutive terms is the same, itβs an arithmetic sequence. Letβs look at another example.

Write the 10th term in the
arithmetic sequence 23, 19, 15, 11, 7.

Weβve been given terms one through
five. Weβre not worried about terms six
through nine. We only need to find the 10th
term. The first thing we should notice is
that the sequence is decreasing. Each term is four less than the
previous term. And so the common difference is
negative four. Letβs consider two different ways
to work out the 10th term.

We could solve for the sixth,
seventh, eighth, and ninth term and then find the 10th term. We would need to subtract four from
each of the previous terms. Or we could start at the fifth
term, seven, and then realize that we need to add five lots of negative four to get
to the 10th term. In the first way, seven minus four
is three, three minus four is negative one, negative one minus four is negative
five, negative five minus four is negative nine, and negative nine minus four is
negative 13.

Using the other method, we would
start at seven, and we would add five lots of negative four. So seven plus five times negative
four equals seven minus 20. Five times negative four is
negative 20. And seven minus 20 is negative
13. Both methods show the 10th term is
negative 13.

But what if our question had asked
us to write the 250th term of the sequence? We wouldnβt want to work out all
250 of the terms. So letβs consider a way to find a
general formula that will give us any term in the sequence.

If we have a sequence three, seven,
11, 15, the common difference is positive four. Weβre looking for the general
form. We can let the letter π stand for
the term number. That is the position in our
sequence. We can use the notation π‘ and
parentheses π to represent the πth term. When π is one, π‘ one is
three. When π is two, π‘ of two equals
seven, and so on. If we look closely, we see every
time our position in the sequence goes up by one, the value of the term in the
sequence goes up by four. The value of the term is going up
four times faster than the value of the position. And so we know that part of our
formula will have plus four π.

If positive four π was our general
formula, we would say π‘ of π equals four π. When π equals one in the first
position, π‘ of π would equal four. When π equals two, π‘ of π would
be equal to eight. When π equals three, π‘ of π
would be equal to 12. And when π equals four, π‘ of π
would be equal to 16. But this formula is not exactly
right. When π equals one, we should have
three and not four. When π equals two, we should have
seven and not eight. When π equals three, we should
have 11 and not 12. And when π is four, we should have
15 and not 16. All of these values were one
off. Four minus one is three, eight
minus one is seven, 12 minus one is 11, and 16 minus one is 15. And so we can add minus one to our
π‘ of π formula so that the new general formula is π‘ of π is equal to positive
four π minus one, but we donβt have to include the positive sign.

The formula for finding the πth
term of this sequence will be π‘ of π equals four π minus one. If we use this form to calculate
the 250th term, four times 250 is 1000, 1000 minus one is 999. The 250th term in this sequence is
999. We should also point out that there
are a few different notations for π‘ of π. You might use π‘ sub π or even π
sub π. All three of our ways to represent
the πth term.

Letβs try that method on a new
problem.

Find the 81st term in the sequence
107, 99, 91, 83.

Weβve been given the first four
terms in a sequence. Weβll let π be the position number
or the term number and π‘ of π is the value of the term in the sequence. We recognize that the sequence is
going down by eight; our common difference is negative eight. To make a general formula for this
sequence, we would need negative eight π. If we plug in our π values one
through four, we would end up with the values negative eight, negative 16, negative
24, negative 32. But when π equals one, our value
is 107 not negative eight. When π equals two, we need 99 not
negative 16. For the third term, we need 91 and
not negative 24. And for the fourth term, we need 83
not negative 32.

If we add 115 to negative eight, we
would get 107. If we add 115 to negative 16, we
would get 99. And so we need to add 115 to the
negative eight π. A general formula for this sequence
would look like this. π‘ of π is equal to negative eight
π plus 115. To solve for the 81st term in the
sequence, we would need π‘ of 81, which is equal to negative eight times 81 plus
115. And thatβs negative 533. The 81st term in this sequence is
negative 533.

Finding the general form with this
method still requires a lot of calculation, so letβs consider another way to do
that. If we take our π and π‘ of π
values and plot them along an π₯- and π¦-axis, weβll start with the table that looks
something like this. Plotting those points would give us
something that looks like this. π can only take integer values,
one, two, three, four, five, and so on. Because our sequence has terms one,
two, three, four, and so on. It doesnβt make sense to talk about
the 3.75 term, but letβs consider what would happen if we connected these points
with a line.

The line would cross the π¦-axis at
115. This is because when we increase
our π₯-value by one, the π¦-value decreases by eight. And this means if we do the
opposite β if we decrease our π₯-value by one β we need to increase our π¦-value by
eight. If we move to the left from π₯
equals one, we get the place where π₯ equals zero. And we need to increase by
eight. And we see that when π₯ equals
zero, π¦ equals 115. If we were talking about this like
a line, we would say it has a slope of negative eight and a π¦-intercept of 115. We would say that π¦ equals
negative eight π₯ plus 115.

And we can use this to help us find
the general form in an arithmetic sequence. The common difference is negative
eight. The common difference told us the
slope of the line. And the π¦-intercept told us the
number we needed to add. This matches the form we found
previously, π‘ of π equals negative eight π plus 115. Letβs try and use this method on a
different problem.

Find the formula for the πth term
of the arithmetic sequence 6.8, 7.9, 9.0, 10.1.

Weβre given the first four terms in
an arithmetic sequence. Weβll let the term number be equal
to π and π‘ of π is the corresponding value of the term in the sequence. To get from term one to term two,
we add 1.1. To get from term two to three, we
add 1.1. And term three plus 1.1 equals term
four. The common difference is positive
1.1. What we want to do now is find the
value of our zero term.

When we move along consecutively
from left to right, weβre adding 1.1. If we want to move in the opposite
direction, we would be subtracting 1.1. To find our zeroth term, we need to
subtract 1.1 from our first term. 6.8 minus 1.1 equals 5.7. The formula for the πth term of
this sequence equals 1.1 times π plus 5.7.

Letβs summarize what weβve just
shown. We could say that the formula for
the πth term of an arithmetic sequence looks like this. π‘ of π equals π times π plus
π. The constant π is the value we
multiply by the term number, the common difference, which can be positive or
negative. And the π value is the constant
that is the zeroth term of the sequence. If we see formulas with squares,
square roots, or powers of π, the sequence is not an arithmetic sequence. All of these are nonlinear
equations and, therefore, cannot represent an arithmetic sequence.

Letβs consider a final type of
question you might see when dealing with arithmetic sequences.

Does 117 belong to the arithmetic
sequence five, 18, 31, 44?

Weβre given the first four terms in
an arithmetic sequence. Weβll let π be the term number and
π‘ of π is the value of the term in the sequence. The first thing we should check
here is the common difference. To get from term to term here,
weβre adding 13. And the common difference is then
positive 13. At this point, itβs good to
remember π‘ of π equals π times π plus π, where π is the common difference and
π is the zeroth term. And this means the next thing we
need to find is what the zero term is.

In this sequence, when weβre moving
to the right, weβre adding 13. And that means to move to the left,
we need to subtract 13. Five minus 13 is negative
eight. And negative eight is the zero
term. We now have enough information to
write the general form for the πth term of this sequence. That will be 13π minus eight, a
common difference of positive 13 and a zeroth term of negative eight.

In order to find out if 117 belongs
to this sequence, we plug 117 in for π‘ of π. If 117 is part of this sequence, it
will have an integer value for π. So we solve this equation for π
and see if itβs an integer. To do that, we first add eight to
both sides. This gives us 125 is equal to
13π. And so we divide both sides of the
equation by 13. 125 divided by 13 is equal to
9.6153 continuing or, as a fraction, nine and eight thirteenths. Because 9.6153 continuing is not an
integer, 117 is not found in this sequence. When dealing with arithmetic
sequences and this general form, the π-value must be an integer.

To summarize what we know, an
arithmetic sequence is a sequence in which the difference between any two
consecutive terms is always the same. The common difference is the
difference between any two consecutive terms in an arithmetic sequence. And we can use the general form π‘
of π equals ππ plus π, where π equals the positive or negative common
difference and π equals the zeroth term.