### Video Transcript

The πth term of a geometric
sequence is π sub π, first term is π sub one, and the sum of the first π terms
is π sub π. Find the first three terms and the
sum to β, π sub β, of the infinite geometric sequence with π sub two minus π sub
six equals 33 and π sub four equals 132.

A geometric sequence is one in
which the ratio between consecutive terms is constant. In other words, if we know any term
in the sequence, we always get to the next term in the same way by multiplying by
the same value, which we call the common ratio and denote using the letter π. Weβve been given various pieces of
information about this sequence. The difference between the second
and sixth terms is 33, and the sum of the first four terms is 132. Letβs recall some of the key
formulae that weβll need in this question.

First, the general term of a
geometric sequence, π sub π, is given by π sub one multiplied by π to the power
of π minus one. We take the first term and multiply
it by the common ratio π π minus one times. Secondly, the sum of the first π
terms of a geometric sequence is given by π sub π is equal to π sub one
multiplied by one minus π to the power of π over one minus π. And finally, if the absolute value
of the common ratio π is less than one, then the infinite geometric sequence is
convergent. And we can find the sum to β using
the formula π sub β is equal to π sub one over one minus π. Letβs use these formulae and the
information given in the question to form some equations.

For the second term, the value of
π is two. So the second term π sub two is
equal to π sub one multiplied by π to the power of two minus one, which is simply
π to the first power, or π. For the sixth term, the value of π
is six. So we have π sub six is equal to
π sub one multiplied by π to the power of five. Weβre told that π sub two minus π
sub six is equal to 33. So we have the equation π sub one
π minus π sub one π to the fifth power is equal to 33. We can factor this equation on the
left-hand side by π sub one π to give π sub one π multiplied by one minus π to
the fourth power is equal to 33. And this is our first equation.

Next, weβre told that the sum of
the first four terms π sub four is equal to 132. So substituting π equals four into
the formula for the sum of the first π terms of a geometric sequence gives π sub
one multiplied by one minus π to the fourth power over one minus π is equal to
132. This is our second equation. Now, looking at the two equations
weβve written, we note that both equations contain factors of π sub one multiplied
by one minus π to the fourth power. We can rearrange the first equation
to give π sub one multiplied by one minus π to the fourth power is equal to 33
over π and rearrange the second to give π sub one multiplied by one minus π to
the fourth power is equal to 132 multiplied by one minus π.

As the left-hand sides of these two
equations are equal, we can then equate the right-hand sides, which will give an
equation in π only. 33 over π is equal to 132
multiplied by one minus π. We now look to solve this equation
for π. The first step is to multiply both
sides by π, giving 33 equals 132π multiplied by one minus π. Next, we distribute the parentheses
on the right-hand side, giving 33 is equal to 132π minus 132π squared. And then we can collect all the
terms on the left-hand side of the equation by adding 132π squared and subtracting
132π from both sides to give 132π squared minus 132π plus 33 is equal to
zero.

We now see that we have a quadratic
equation in π. In fact this equation can be
simplified quite significantly because all three coefficients are multiples of
33. Dividing through by 33 then gives
the simplified quadratic four π squared minus four π plus one is equal to
zero. Now, in fact this quadratic
equation can be solved by factoring. And it is a perfect square. Four π squared minus four π plus
one is equal to two π minus one multiplied by two π minus one, or simply two π
minus one squared. We then solve for π by setting the
expression inside the parentheses equal to zero because if its square is zero, it
must itself be zero. So we have two π minus one is
equal to zero. And then adding one to both sides
of this equation and dividing by two gives π is equal to one-half.

So we found the value of π, the
common ratio of this geometric sequence. And we should be encouraged at this
point because weβre later asked to find the sum to β of this geometric sequence. And in order to be able to do this,
we need the absolute value of π to be less than one, which is the case when π is
equal to one-half. Next, we need to find the value of
π sub one, the first term in the sequence, which we can do by substituting π
equals one-half into equation one. Doing so gives π sub one
multiplied by a half multiplied by one minus a half to the fourth power is equal to
33. A half to the fourth power is one
over 16, and then one minus one over 16 is 15 over 16. We then multiply 15 over 16 by
one-half, which gives 15 over 32. So we have π sub one multiplied by
15 over 32 is equal to 33.

To find the value of π sub one, we
need to multiply both sides of this equation by the reciprocal of 15 over 32, which
is 32 over 15. And we have that π sub one is
equal to 33 multiplied by 32 over 15, which simplifies to 352 over five. So we found the first term in this
sequence. We also need to find the next two
terms and then the sum to β. So letβs create some space to do
this. Remember, to get from one term to
the next in a geometric sequence, we multiply by the common ratio. So if the first term is 352 over
five, the second term is 352 over five multiplied by one-half, which is 176 over
five. The next term, the third term, is
176 over five multiplied by one-half, which is 88 over five.

So we found the first three terms,
and now we need to calculate the sum to β. Remember, the formula for this is
π sub one over one minus π, where the absolute value of π must be strictly less
than one. So we have 352 over five multiplied
by one over one minus a half. Now one minus a half is a half, and
one divided by a half is two. So this simplifies to 352 over five
multiplied by two, which is 704 over five. So weβve completed the problem. By first calculating the common
ratio and the first term in the sequence, we found that the first three terms are
352 over five, 176 over five, and 88 over five. And the sum to β of this infinite
geometric sequence is 704 over five.