Video Transcript
In this video, we’re gonna use your
skills of algebraic manipulation in order to solve some geometric sequence
problems. First, let’s do a little warm up
with this question.
The fourth and fifth terms in a
geometric sequence are 125 and 625 respectively. Find the first, second, third,
and 10th terms of the sequence.
Right, first let’s just define
some terms. I’m gonna use 𝑎 one to
represent the first term in the sequence, 𝑎 𝑛 to represent the 𝑛th term. So 𝑎 two is the second term,
𝑎 three is the third term, and so on. And then I’m gonna use the
letter 𝑟 to represent the common ratio between terms. So we’ve been given the fourth
and fifth terms, so 𝑎 four is 125 and 𝑎 five is 625. And remember, to get one term,
we need to multiply the previous term by the common ratio. So 𝑎 five for example is 𝑎
four times the common ratio. So that means, in our case,
that 625 is 125 times 𝑟. And dividing both sides of that
equation by 125, I get 625 over 125 is equal to 𝑟. In other words, in this case,
𝑟 is equal to five.
So just remember then that to
work out the value of 𝑟, the common ratio, we take the value of one term and we
divide it by the value of the immediately preceding term. So, so far, we’ve worked out
that the value of the common ratio is five. So to work out the value of 𝑎
four, we just take 𝑎 three and multiply it by 𝑟. Well, we know the value of 𝑎
four; we know the value of 𝑟. So we can now work out the
value of 𝑎 three, the third term. 125 equals 𝑎 three times
five. So dividing both of those by
five means that the third term is equal to 25. And the third term is equal to
the second term times 𝑟.
Well, now we know the third
term, and we know the value of 𝑟. So we can work out the value of
the second term. 25 equals 𝑎 two times
five. So, again, we can divide both
sides by five. And we now know that the value
of the second term is five. Also, the value of the second
term is the value of the first term times the common ratio. Well, the value of the second
term is five. The value of the common ratio
is five. So I’ve got five is equal to
the first term times five. Dividing both sides by five
tells me that the first term has a value of one. So I’ve worked out the value of
the first, second, and third terms.
To work out the value of the
10th, I could just roll this procedure forwards. But I’m gonna work out the
value of the 𝑛th term, so a formula for the 𝑛th term. And then apply that to the
value 10. Now, the general formula for
the 𝑛th term of a geometric sequence is that the 𝑛th term is equal to the
first term times the common ratio to the power of 𝑛 minus one. Well, I know that the first
term has a value of one. I know that the common ratio is
five. So I can now complete this
formula. 𝑛th term is one times five to
the power of 𝑛 minus one. Well, one times that, I don’t
really need to write one times that. So I can make that a little bit
more efficient. So my 𝑛th term formula is 𝑎
𝑛, the 𝑛th term, is equal to five to the power of 𝑛 minus one.
Now, I can use that formula to
calculate the value of the 10th term. So I just put in a value of 𝑛
equals 10. So my 10th term is gonna be
five to the power of 10 minus one. So that’s five to the power of
nine. And when I put that into my
calculator, I get 1953125. So just checking back to the
question to make sure we’ve answered it. We had to find the value of the
first, second, third, and 10th terms of the sequence. Well, the first, second, and
third terms are one, five, and 25. And the 10th term, as we just
said, is 1953125.
Okay, now we’re warmed up. Let’s do another question that
actually genuinely involves a bit more algebra.
A geometric sequence that
consists of only positive terms has the first three terms two, 𝑥 minus two, and
𝑥 plus 10. Find the possible values of 𝑥
and the general formula for the 𝑛th term of the sequence.
So we’re told that the first
term is two, the second term is 𝑥 minus two, and the third term is 𝑥 plus
10. Now, we weren’t told what the
common ratio was. But we do know that the common
ratio 𝑟 is simply one term divided by the immediately preceding term. So it could be 𝑎 two divided
by 𝑎 one or 𝑎 three divided by a two, for example. And the consequence of that is
that 𝑎 two divided by 𝑎 one must give us the same answer as 𝑎 three divided
by 𝑎 two. Now, we’ve got expressions for
𝑎 one, 𝑎 two, and 𝑎 three. So let’s put them into one
great big equation.
So that means 𝑥 minus two over
two is equal to 𝑥 plus 10 over 𝑥 minus two. Now, if I multiplied both sides
of that equation by two, then the two cancels from the left-hand side, just
giving me 𝑥 minus two. And then, I’ve got an extra two
on the numerator on the right-hand side. Now, I’m gonna multiply both
sides of my equation by 𝑥 minus two to try to eliminate the denominator from
the right-hand side. So on the left-hand side, I had
my 𝑥 minus two. And then, I multiplied that by
𝑥 minus two. And on the right-hand side, the
𝑥 minus two that I am multiplying by cancels the denominator. So it just leaves me with two
lots of 𝑥 plus 10. So we’re just making a little
bit of space for ourselves.
We’ve got 𝑥 minus two times 𝑥
minus two is equal to two times 𝑥 plus 10. So I’m gonna multiply those out
on the left-hand side, multiplying each term in the second bracket by each term
in the first bracket. And distributing the two across
the brackets or parentheses on the right-hand side. So on the left, I’ve got 𝑥
squared minus two 𝑥 minus another two 𝑥 plus four. And on the right-hand side, I
got two times 𝑥 is two 𝑥 plus two times 10 is 20. So here, I’ve got minus two 𝑥
minus another two 𝑥, so I can combine those to make minus four 𝑥. So I’ve got a quadratic
expression, so what I’m gonna do is try and solve that. So I’ve got to make that equal
to zero. So I’m gonna subtract two 𝑥
from both sides and then subtract 20, giving me a quadratic equals to zero. Which, hopefully, I’ll be able
to factor and then find out the values of 𝑥.
So, first of all, let’s
subtract two 𝑥. So on the left-hand side, we’ve
got 𝑥 squared minus four 𝑥 minus another two 𝑥 is minus six 𝑥. And then we still got our plus
four. And on the right-hand side,
we’re taking away that two 𝑥, which just leaves us with 20. So now, let’s subtract 20. So on the left-hand side, I’ve
got 𝑥 squared minus six 𝑥 plus four take away 20 is minus 16, negative 16. And on the right-hand side, if
I take away 20, I’ve just got nothing. So now, I’ve got a quadratic
which is equal to zero. If I can factor that quadratic,
I can find out the values of 𝑥. And luckily, it’s quite easy to
factor. 𝑥 plus two times 𝑥 minus
eight. So those two things multiplied
together are equal to zero.
And the only way you can get an
answer of zero if you multiply two things together is if one of them is
zero. So either 𝑥 plus two is zero
or 𝑥 minus eight is zero. And if 𝑥 plus two is equal to
zero, then 𝑥 will be equal to negative two. And to make 𝑥 minus eight
equal to zero, 𝑥 would need to be equal to eight. So we’ve got two possible
values of 𝑥. Now remember, the question said
that the geometric sequence consists of only positive terms. So these terms have to be
positive. We also know that the first
term is two. The second term is 𝑥 minus
two. And the third term is 𝑥 plus
10. So if 𝑥 was equal to negative
two, the second term would be negative two take away two, be negative four. So that would break the
rule. So that can’t actually be a
solution.
So the only value of 𝑥 that
satisfies the instructions in the question, if you like, is 𝑥 equals eight. And if 𝑥 equals eight, I can
now write out the first three terms of my sequence properly. Well, 𝑎 one is still two. 𝑎 two is 𝑥 minus two, so
that’s eight minus two which is six. And 𝑎 three is 𝑥 plus 10. So that’s eight plus 10 is
18. So the first three terms are
two, six, and 18. And I can work out the common
ratio 𝑟 just by doing the second term divided by the first term. So that’s six divided by
two. So the common ratio is
three. And the general formula for the
𝑛th term of any sequence is 𝑎 𝑛, the 𝑛th term, is equal to the first term,
𝑎 one, times the common ratio, 𝑟, to the power of 𝑛 minus one. Now, we’ve got values for 𝑎
one and 𝑟. My 𝑛th term formula becomes
the 𝑛th term, 𝑎 𝑛, is equal to two times three to the power of 𝑛 minus
one.
And strictly, you don’t need to
put the parentheses around the three. But it just makes it absolutely
clear that it’s only the three that is to the power of 𝑛 minus one and not the
two as well. So we’re just going back and
checking we’ve answered everything. Find the possible values of
𝑥. Well, there is only one
possible value of 𝑥, if the sequence is only gonna consist of positive
terms. And we wanted the general
formula for the 𝑛th term of the sequence. And we’ve got that here.
Okay, looks like we’re done.
