The third term in an arithmetic
sequence is two and the sixth term is 11. If the first term is 𝑎 one, what
is an equation for the 𝑛th term of this sequence?
So in order to actually solve this
problem, there’s a couple of methods we can use. So I’m actually gonna give an
example of each. So the first thing to bear in mind
is that actually we’re looking at an arithmetic sequence. And what we have when we’re dealing
with an arithmetic sequence is a general rule for any term. And that general rule is that 𝑎
𝑛, so the 𝑛th term, is equal to the first term plus 𝑛 minus one — and 𝑛 is the
term position — multiplied by 𝑑, which is our common difference.
Okay, so now we actually have this
general rule, let’s actually use this to help us find out what the actual 𝑛th term
of our sequence is going to be. So if we take a look at the
question, we have two bits of information that are very important. We have the third term, which is
equal to two, and the sixth term, which is equal to 11.
So what we can actually do is use
these two bits of information to actually form two simultaneous equations, which can
actually help us find our 𝑎 and 𝑑, so our first term and our common
difference. So first of all, if we use the fact
that the third term is equal to two, we can say that two is equal to 𝑎 one, so our
first term, plus three minus one 𝑑, and that’s three minus one because actually
that was our term number, which is three, and then minus one.
So we can actually just rewrite
this as two is equal to 𝑎 one, so our first term, plus two 𝑑, and that’s cause
three minus one is two. And then we can label this equation
one because this is our first simultaneous equation. And then using the second bit of
information, which is that the sixth term is equal to 11, we can say that 11 is
equal to 𝑎 one plus six minus one 𝑑. So therefore, we can write this as
11 is equal to 𝑎 one plus five 𝑑, and that’s because we have six minus one which
is five. Great! And we’ve also labeled this
So now to enable us to find our
𝑛th term, what we’ll need to know is our 𝑎 one, so our first term, and our 𝑑,
which is our common difference. So what we’re gonna do is actually
subtract equation one from equation two to eliminate our 𝑎 one. And when we do this, we get nine is
equal to three 𝑑, and that’s because you had 11 minus two, which is nine, and then
you had 𝑎 one minus 𝑎 one, which is zero, and then five 𝑑 minus two 𝑑, which is
And then if we actually divide both
sides of the equation by three, we’re gonna get three is equal to 𝑑, so we’ve found
the value of 𝑑. So the common difference is going
to be three. So now what we can do is actually
substitute 𝑑 equals three into equation one. You could actually substitute it
into either equation, but we’re gonna choose equation one for this case to find 𝑎
So if we do that, we get two is
equal to 𝑎 one plus two multiplied by three. So we’re gonna get two is equal to
𝑎 one plus six. So then if we subtract six from
each side of the equation, we get negative four is equal to 𝑎 one. So therefore, we’ve actually found
our first term.
So now what we can do is actually
write our 𝑛th term. So we can say 𝑎 𝑛, so our 𝑛th
term, is equal to negative four, because this is our 𝑎 one value, then plus three
multiplied by 𝑛 minus one, and that’s because our 𝑑 value, our common difference,
So now we can expand the
parentheses. And when we do that, we’re gonna
get our 𝑛th term is equal to negative four plus three 𝑛 because we had three
multiplied by 𝑛, which gives us positive three 𝑛 and then minus three, because we
had positive three multiplied by negative one. So therefore, we get 𝑎 𝑛, so our
𝑛th term, is equal to three 𝑛 minus seven, and that’s because we had negative four
minus three, which gave us negative seven.
Okay, great! So we’ve solved the problem. But what I said is that I’ll
actually show you another method you could have used to actually find this. So we’re gonna do that now just to
check our answer. Okay, so to use this method, what
I’ve actually done is that I’ve written out our term numbers: one, two, three, four,
five, six. And I’ve actually put in the
information we’ve got.
So we know that the third term is
equal to two and the sixth term is equal to 11. And what we can actually see is
that there are actually three jumps to get from three to six. So we get from three to four, four
to five, and five to six. And we know the difference between
two and 11 is actually nine, because you have to add nine onto two to get to 11.
So therefore, what this will give
us is our common difference of positive three. So and if we’re trying to find the
𝑛th term, we know that it’s gonna be 𝑎 𝑛, so our 𝑛th term, is equal to three
𝑛. But then we need to work out, well,
it’s gonna be three 𝑛, what three 𝑛 plus anything three 𝑛 minus anything.
So to work out the next part of our
𝑛th term, we can actually look at the term we had, which is a third term, which
equals two. Well, we can see that three
multiplied by three is gonna be equal to nine, so what’d that be is because it’s
three 𝑛 it’s three multiplied by our term number, which is three, so three by
three, which is nine. But we know that the actual term
value is gonna be two because we know the third term is equal to two.
So what do we need to do to get
from nine to two? Well, we’re needing to actually
subtract seven. So now what we use is we use the
negative seven to actually form the second part of our 𝑛th term. So we can say that the 𝑛th term is
equal to three 𝑛 minus seven. So then if we check back to
previous method, it’s exactly the same.
So great! What we’ve actually done is shown
you how to use two methods. And we’ve also checked our
answer. So we can say that if the third
term of an arithmetic sequence is two and the sixth term is 11 and the first term is
𝑎 one, the equation for the 𝑛th term of that sequence is going to be equal to
three 𝑛 minus seven.