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Let's write an arithmetic
sequence in general terms.

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So we can start
with some number a.

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And then we can
keep adding d to it.

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And that number
that we keep adding,

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which could be a positive
or a negative number,

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we call our common difference.

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So the second term in our
sequence will be a plus d.

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The third term in our
sequence will be a plus 2d.

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So we keep adding d all
the way to the n-th term

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in our sequence.

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And you already see here
that in our first term,

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we added d zero times.

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Our second term,
we added d once.

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In our third term,
we added d twice.

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So you see, whatever the
index of the term is,

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we're adding d one less
than that many times.

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So if we go all the
way to the n-th term,

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we're going to add d
one less than n times.

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So it's going to be
n minus 1 times d.

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Fair enough.

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And let me write that.

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This right over here
is our n-th term.

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Now what I want to do
is think about what

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the sum of this arithmetic
sequence would be.

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And the sum of an
arithmetic sequence

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we call an arithmetic series.

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So let me write that in yellow.

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Color changing is
sometimes difficult.

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So the arithmetic
series is just the sum

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of an arithmetic sequence.

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So let's call my
arithmetic series s sub n.

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And let's say it's going to
be the sum of these terms,

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so it's going to be a plus
d, plus a plus 2d, plus all

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the way to adding the n-th term,
which is a plus n minus 1 times

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d.

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Now I'm going to
do the same trick

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that I did when I did the most
basic arithmetic sequence.

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I'm going to add this
to itself, but I'm

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going to swap the order
in which I write this sum.

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So s sub n I can
write as this, but I'm

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going to write it
in reverse order.

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I'm going to write
the last term first.

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So the n-th term is a
plus n minus 1 times d.

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Then the second to
last term is going

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to be a plus n minus 2 times d.

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The third to last is going to
be a plus n minus 3 times d.

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And we're going to
go all the way down

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to the first term,
which is just a.

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Now let's add these
two equations.

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We are going to get, on the
left hand side, s sub n plus s

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sub n.

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You're going to get
2 times s sub n.

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Well, what's the sum of
these two first terms

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right over here?

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I'm going to have a plus
a plus n minus 1 times d.

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So it's going to be 2a
plus n minus 1 times d.

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Now let's add both of
these second terms.

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So if I were to add both
of these second terms,

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what do I get it?

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I'm going to get 2a plus 2a.

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And what's d plus
n minus 2 times d?

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So you could view
it several ways.

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Let me write this over here.

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What is d plus n
minus 2 times d?

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Well, this is just
the same thing

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as 1d plus n minus 2 times d.

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And so you could just
add the coefficients.

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So this is going to be n
minus 2 plus 1 times d, which

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is equal to n minus 1 times d.

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So the second term also becomes
2a plus n minus 1 times d.

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Now let's add the third term.

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I'll do it in green.

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The third terms, I should say.

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And I think you're going
to see a pattern here.

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It's 2a plus 2a.

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And if I have 2 plus n minus 3
of something and then I add 2,

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I'm going to have n minus
one of that something.

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So plus n minus 1 times d.

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And you're going to keep
doing that all the way

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until your n-th pair
of terms, all the way

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until you add these two
characters over here, which

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is just 2a plus n
minus 1 times d.

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So you have this
2a plus n minus 1 d

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being added over and over again.

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And how many times
are you doing that?

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Well, you had n
pairs of terms when

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you were adding
these two equations.

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In each of them,
you had n terms.

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This is the first term,
this is the second term,

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this is the third term, all
the way to the n-th term.

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So I can rewrite 2
times the sum 2 times

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s sub n is going to be
n times this quantity.

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It's going to be n times
2a plus n minus 1 times d.

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And then if we want
to solve for s sub n,

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you just divide both sides by 2.

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And you get s sub n is equal
to, and we get ourselves

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a little bit of
a drum roll here,

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n times 2a plus n
minus 1 times d.

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All of that over 2.

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Now, we've come up
with a general formula,

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just a function of
what our first term is,

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what our common difference
is, and how many terms we're

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adding up.

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And so this is the generalized
sum of an arithmetic sequence,

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which we call an
arithmetic series.

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But now, let's ask
ourselves this question.

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This is hard to remember.

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The n times 2a plus n
minus 1 times d over 2.

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But in the last video, when I
did a more concrete example,

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I said well, it looks like the
sum of an arithmetic sequence

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could be written as
perhaps the average

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of the first term a1 plus an.

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The average of the first
term and the last term

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times the number of
terms that you have.

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So is this actually the case?

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Do these two things gel?

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Because this is very
easy to remember--

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the average of the first and
the last terms multiplied

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by the number of terms
you had and actually makes

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intuitive sense,
because you're just

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increasing by the same
amount every time.

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So let's just average the
first and the last term

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and then multiply times the
number of terms we have.

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Well, all we have to do is
rewrite this a little bit

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to see that it is indeed
the exact same thing as this

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over here.

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So all we have to do
is break out the a.

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So let me rewrite it.

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So, this could be
rewritten as s sub

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n is equal to n times a plus
a plus n minus 1 times d.

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I just broke up this
2a into an a plus a.

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All of that over 2.

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And you see, based on how
we defined this thing,

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our first term a1 is a.

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And then our last term, a sub
n, is a plus n minus 1 times d.

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So this whole business
right over here

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really is the average of
the first and last terms.

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I got my first term,
adding it to my last term,

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dividing it by 2.

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And then I'm multiplying by
the number of terms we have.

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And that's true for any
arithmetic sequence,

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as we've just shown here.