### Video Transcript

Write down the first term and the
common ratio for the following geometric sequence: 10, negative five, five over two,
negative five over four, and so on.

Well clearly, the first term is
equal to 10. So π one is equal to 10; that bit
was quite easy. And the common ratio is what do we
multiply each term by to get the next term. So Iβm just gonna label all of my
terms π one, π two, π three, and π four, and so on. And then Iβm just going to write a
little formula for how do I get from one term to the next term. Well, if I multiply the first term
by the common ratio, π, I get the second term. If I multiply the second term by
the common ratio, π, I get the third term. If I multiply the third term by the
common ratio, π, I get the fourth term, and so on and so on and so on. So looking at that first equation,
if I divide both sides of the equation by π one, I get that π is equal to π two
over π one. Now, if I divide both sides of the
second equation by π two, I get that π is equal to π three over π two and
similarly for the third equation.

So to work out the value of π, I
just take the value of one term and divide it by the value of the previous term. Now remember in a geometric
sequence, itβs a common ratio. So it doesnβt matter whether I take
the second and the first or the third and the second or the fourth and the
third. As long as I take two consecutive
terms, I will always find the same answer for π. Well looking at these numbers here,
the easiest pair to use is gonna be π one and π two. So π is equal to π two divided by
π one. π two is negative five and π one
is ten. So the common ratio is negative
five divided by 10 and that simplifies to negative a half.

So to get from each term to the
next term, I have to multiply by negative a half. 10 times negative a half is minus
five, negative five times negative a half is five over two, and so on. So these two facts here, π one
equals 10 and π equals negative a half, uniquely defines this sequence. When we know this, we can generate
all the terms in the sequence if weβre prepared to put in enough time and doing
enough multiplying.