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## Finite arithmetic series

Current time:0:00Total duration:7:47

# Arithmetic series formula

## Video transcript

Let's write an arithmetic
sequence in general terms. So we can start
with some number a. And then we can
keep adding d to it. And that number
that we keep adding, which could be a positive
or a negative number, we call our common difference. So the second term in our
sequence will be a plus d. The third term in our
sequence will be a plus 2d. So we keep adding d all
the way to the n-th term in our sequence. And you already see here
that in our first term, we added d zero times. Our second term,
we added d once. In our third term,
we added d twice. So you see, whatever the
index of the term is, we're adding d one less
than that many times. So if we go all the
way to the n-th term, we're going to add d
one less than n times. So it's going to be
n minus 1 times d. Fair enough. And let me write that. This right over here
is our n-th term. Now what I want to do
is think about what the sum of this arithmetic
sequence would be. And the sum of an
arithmetic sequence we call an arithmetic series. So let me write that in yellow. Color changing is
sometimes difficult. So the arithmetic
series is just the sum of an arithmetic sequence. So let's call my
arithmetic series s sub n. And let's say it's going to
be the sum of these terms, so it's going to be a plus
d, plus a plus 2d, plus all the way to adding the n-th term,
which is a plus n minus 1 times d. Now I'm going to
do the same trick that I did when I did the most
basic arithmetic sequence. I'm going to add this
to itself, but I'm going to swap the order
in which I write this sum. So s sub n I can
write as this, but I'm going to write it
in reverse order. I'm going to write
the last term first. So the n-th term is a
plus n minus 1 times d. Then the second to
last term is going to be a plus n minus 2 times d. The third to last is going to
be a plus n minus 3 times d. And we're going to
go all the way down to the first term,
which is just a. Now let's add these
two equations. We are going to get, on the
left hand side, s sub n plus s sub n. You're going to get
2 times s sub n. Well, what's the sum of
these two first terms right over here? I'm going to have a plus
a plus n minus 1 times d. So it's going to be 2a
plus n minus 1 times d. Now let's add both of
these second terms. So if I were to add both
of these second terms, what do I get it? I'm going to get 2a plus 2a. And what's d plus
n minus 2 times d? So you could view
it several ways. Let me write this over here. What is d plus n
minus 2 times d? Well, this is just
the same thing as 1d plus n minus 2 times d. And so you could just
add the coefficients. So this is going to be n
minus 2 plus 1 times d, which is equal to n minus 1 times d. So the second term also becomes
2a plus n minus 1 times d. Now let's add the third term. I'll do it in green. The third terms, I should say. And I think you're going
to see a pattern here. It's 2a plus 2a. And if I have 2 plus n minus 3
of something and then I add 2, I'm going to have n minus
one of that something. So plus n minus 1 times d. And you're going to keep
doing that all the way until your n-th pair
of terms, all the way until you add these two
characters over here, which is just 2a plus n
minus 1 times d. So you have this
2a plus n minus 1 d being added over and over again. And how many times
are you doing that? Well, you had n
pairs of terms when you were adding
these two equations. In each of them,
you had n terms. This is the first term,
this is the second term, this is the third term, all
the way to the n-th term. So I can rewrite 2
times the sum 2 times s sub n is going to be
n times this quantity. It's going to be n times
2a plus n minus 1 times d. And then if we want
to solve for s sub n, you just divide both sides by 2. And you get s sub n is equal
to, and we get ourselves a little bit of
a drum roll here, n times 2a plus n
minus 1 times d. All of that over 2. Now, we've come up
with a general formula, just a function of
what our first term is, what our common difference
is, and how many terms we're adding up. And so this is the generalized
sum of an arithmetic sequence, which we call an
arithmetic series. But now, let's ask
ourselves this question. This is hard to remember. The n times 2a plus n
minus 1 times d over 2. But in the last video, when I
did a more concrete example, I said well, it looks like the
sum of an arithmetic sequence could be written as
perhaps the average of the first term a1 plus an. The average of the first
term and the last term times the number of
terms that you have. So is this actually the case? Do these two things gel? Because this is very
easy to remember-- the average of the first and
the last terms multiplied by the number of terms
you had and actually makes intuitive sense,
because you're just increasing by the same
amount every time. So let's just average the
first and the last term and then multiply times the
number of terms we have. Well, all we have to do is
rewrite this a little bit to see that it is indeed
the exact same thing as this over here. So all we have to do
is break out the a. So let me rewrite it. So, this could be
rewritten as s sub n is equal to n times a plus
a plus n minus 1 times d. I just broke up this
2a into an a plus a. All of that over 2. And you see, based on how
we defined this thing, our first term a1 is a. And then our last term, a sub
n, is a plus n minus 1 times d. So this whole business
right over here really is the average of
the first and last terms. I got my first term,
adding it to my last term, dividing it by 2. And then I'm multiplying by
the number of terms we have. And that's true for any
arithmetic sequence, as we've just shown here.