### Video Transcript

Which of the following sequences is
not classified as arithmetic or geometric? Is it (A) one-third, negative
one-third, negative one, negative five over three, and so on? (B) One-half, one-quarter,
one-eighth, one sixteenth, and so on. (C) One-half, one, three over two,
two, and so on. (D) One, one-half, zero, negative
one-half, and so on. Or is it (E) one-half, one-third,
one-quarter, one-fifth, and so on?

To start off with, we should
probably remind ourselves what the terms arithmetic and geometric sequence mean. An arithmetic sequence is one in
which each term can be obtained from a previous term by adding a common
difference. In other words, if we subtract the
πth term in the sequence from the π plus oneth term, where π is any natural
number, the answer will always be a constant π. On the other hand, a geometric
sequence is one where each term can be obtained from a previous term by multiplying
by a common ratio. Mathematically speaking, if we
divide the π plus oneth term by the πth term for any natural number π, we will
always get the constant π.

With each of our options, we have
been given four terms of the sequence, and the rest of the terms can be assumed to
follow the same pattern. Therefore, the easiest way to
classify the options is to verify in each case whether the terms weβve been given
satisfy either of the qualities shown on the right. Let us start by considering option
(A).

First of all, to make things
easier, we can label the terms as π one, π two, π three, and π four. We can then test whether the
sequence is arithmetic by subtracting each term from the next term and seeing
whether the difference is the same. For π two minus π one, we have
negative one-third minus one-third, which is negative two-thirds. For π three minus π two, we have
negative one minus negative one-third, which is the same thing as negative one plus
a third, which is also negative two-thirds. Finally, just to be sure, we
subtract π four from π three, which is negative five-thirds minus negative one,
which is also negative two-thirds. We can see that the difference in
each case is the same. So, we can conclude that the terms
satisfy the requirements for being an arithmetic sequence. So, we can eliminate (A) as an
option.

Next, let us consider (B). We can label the terms π one, π
two, π three, and π four. In exactly the same way as before,
we can consider the differences between successive terms to see whether the sequence
is arithmetic. However, even if we just consider
the first two differences, we can see that they are not the same. In fact, they are getting
smaller. So, let us consider checking
whether the sequence is geometric instead. We do this by taking successive
ratios between the terms and comparing them, or π two over π one, we have
one-quarter over one-half. We can evaluate this by taking the
four into the denominator and two into the numerator, giving us two over four, which
simplifies to a half.

Next, we have π three over π two,
which is one-eighth over one-quarter. Using the same technique as before,
this is four over eight, which is one-half again. We can then confirm that we also
get one-half as the ratio for π four over π three. Since the ratio is the same between
successive terms, this means we have a geometric sequence. So, let us also cross out (B) as an
option.

Now that we are acquainted with the
general procedure, let us consider an abbreviated approach for tackling the
remaining options. For (C), we have the terms
one-half, one, three over two, and two. We can see by inspection that each
term can be obtained from the previous term by adding a half. Since thereβs a common difference
between terms, this means we have an arithmetic sequence. We could also verify this using the
procedure from before by subtracting each term from the subsequent term in the
sequence. So, option (C) is excluded.

Now, let us consider option
(D). Listing out the terms one, a half,
zero, and negative a half, we might notice that this is actually similar to the
previous sequence. The difference is that this time
weβre subtracting a half each time rather than adding a half. However, this is still an
arithmetic sequence since the difference is still constant between successive
terms. So, we can eliminate it.

At this point, we should expect our
final option to be neither an arithmetic nor a geometric sequence, but let us check
this for good measure. We will denote the terms as π one
to π four to make things easier. Let us just take the first two
pairs of terms for a common difference since we do not expect the sequence to be
arithmetic. For π two minus π one, we have a
third minus a half. We can subtract these fractions by
giving them the same denominator. That is, we have two-sixths minus
three-sixths, which is negative one-sixth. But for π three minus π two, we
have a quarter minus a third, which is three twelfths minus four twelfths, which is
negative one twelfth. We do not need to check the third
difference, since it is already clear we do not have a constant difference between
terms.

Let us now check whether the terms
exhibit the geometric property. We first of all have π two over π
one, which is a third over a half. By taking three into the
denominator and two into the numerator, we get two-thirds. For the second ratio, we have
one-quarter over one-third, and this is three-quarters. Since the ratio is not constant
between terms, we cannot have a geometric sequence. In conclusion, since it is the only
sequence that is neither arithmetic nor geometric, the answer is (E).