### Video Transcript

In this video, weβre gonna use your
skills of algebraic manipulation in order to solve some geometric sequence
problems. First, letβs 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 letβs just define some
terms. Iβm gonna use π one to represent
the first term in the sequence, π π to represent the πth term. So π two is the second term, π
three is the third term, and so on. And then Iβm gonna use the letter
π to represent the common ratio between terms. So weβve been given the fourth and
fifth terms, so π four is 125 and π five is 625. And remember, to get one term, we
need to multiply the previous term by the common ratio. So π five for example is π four
times the common ratio. So that means, in our case, that
625 is 125 times π. And dividing both sides of that
equation by 125, I get 625 over 125 is equal to π. In other words, in this case, π is
equal to five.

So just remember then that to work
out the value of π, 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, weβve worked out that
the value of the common ratio is five. So to work out the value of π
four, we just take π three and multiply it by π. Well, we know the value of π four;
we know the value of π. So we can now work out the value of
π three, the third term. 125 equals π 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 π.

Well, now we know the third term,
and we know the value of π. So we can work out the value of the
second term. 25 equals π 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 Iβve 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 Iβve 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 Iβm gonna work out the value of
the πth term, so a formula for the πth term. And then apply that to the value
10. Now, the general formula for the
πth term of a geometric sequence is that the πth term is equal to the first term
times the common ratio to the power of π 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. πth term is one times five to the
power of π minus one. Well, one times that, I donβt
really need to write one times that. So I can make that a little bit
more efficient. So my πth term formula is π π,
the πth term, is equal to five to the power of π minus one.

Now, I can use that formula to
calculate the value of the 10th term. So I just put in a value of π
equals 10. So my 10th term is gonna be five to
the power of 10 minus one. So thatβs 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 weβve 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 weβre warmed up. Letβs 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, π₯ minus two, and π₯ plus
10. Find the possible values of π₯ and
the general formula for the πth term of the sequence.

So weβre told that the first term
is two, the second term is π₯ minus two, and the third term is π₯ plus 10. Now, we werenβt told what the
common ratio was. But we do know that the common
ratio π is simply one term divided by the immediately preceding term. So it could be π two divided by π
one or π three divided by a two, for example. And the consequence of that is that
π two divided by π one must give us the same answer as π three divided by π
two. Now, weβve got expressions for π
one, π two, and π three. So letβs put them into one great
big equation.

So that means π₯ minus two over two
is equal to π₯ plus 10 over π₯ 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
π₯ minus two. And then, Iβve got an extra two on
the numerator on the right-hand side. Now, Iβm gonna multiply both sides
of my equation by π₯ minus two to try to eliminate the denominator from the
right-hand side. So on the left-hand side, I had my
π₯ minus two. And then, I multiplied that by π₯
minus two. And on the right-hand side, the π₯
minus two that I am multiplying by cancels the denominator. So it just leaves me with two lots
of π₯ plus 10. So weβre just making a little bit
of space for ourselves.

Weβve got π₯ minus two times π₯
minus two is equal to two times π₯ plus 10. So Iβm 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, Iβve got π₯ squared
minus two π₯ minus another two π₯ plus four. And on the right-hand side, I got
two times π₯ is two π₯ plus two times 10 is 20. So here, Iβve got minus two π₯
minus another two π₯, so I can combine those to make minus four π₯. So Iβve got a quadratic expression,
so what Iβm gonna do is try and solve that. So Iβve got to make that equal to
zero. So Iβm gonna subtract two π₯ from
both sides and then subtract 20, giving me a quadratic equals to zero. Which, hopefully, Iβll be able to
factor and then find out the values of π₯.

So, first of all, letβs subtract
two π₯. So on the left-hand side, weβve got
π₯ squared minus four π₯ minus another two π₯ is minus six π₯. And then we still got our plus
four. And on the right-hand side, weβre
taking away that two π₯, which just leaves us with 20. So now, letβs subtract 20. So on the left-hand side, Iβve got
π₯ squared minus six π₯ plus four take away 20 is minus 16, negative 16. And on the right-hand side, if I
take away 20, Iβve just got nothing. So now, Iβve got a quadratic which
is equal to zero. If I can factor that quadratic, I
can find out the values of π₯. And luckily, itβs quite easy to
factor. π₯ plus two times π₯ 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 π₯ plus two is zero or π₯
minus eight is zero. And if π₯ plus two is equal to
zero, then π₯ will be equal to negative two. And to make π₯ minus eight equal to
zero, π₯ would need to be equal to eight. So weβve got two possible values of
π₯. 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 π₯ minus
two. And the third term is π₯ plus
10. So if π₯ 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 canβt actually be a
solution.

So the only value of π₯ that
satisfies the instructions in the question, if you like, is π₯ equals eight. And if π₯ equals eight, I can now
write out the first three terms of my sequence properly. Well, π one is still two. π two is π₯ minus two, so thatβs
eight minus two which is six. And π three is π₯ plus 10. So thatβs eight plus 10 is 18. So the first three terms are two,
six, and 18. And I can work out the common ratio
π just by doing the second term divided by the first term. So thatβs six divided by two. So the common ratio is three. And the general formula for the
πth term of any sequence is π π, the πth term, is equal to the first term, π
one, times the common ratio, π, to the power of π minus one. Now, weβve got values for π one
and π. My πth term formula becomes the
πth term, π π, is equal to two times three to the power of π minus one.

And strictly, you donβt need to put
the parentheses around the three. But it just makes it absolutely
clear that itβs only the three that is to the power of π minus one and not the two
as well. So weβre just going back and
checking weβve answered everything. Find the possible values of π₯. Well, there is only one possible
value of π₯, if the sequence is only gonna consist of positive terms. And we wanted the general formula
for the πth term of the sequence. And weβve got that here. Okay, looks like weβre done.