Video Transcript
Find the order of the term sixteen
thirty-ninth in the geometric sequence one over 156, negative one over 78, one over
39.
If we start by listing out this
geometric sequence, we know that sixteen thirty-ninth falls somewhere in the
sequence. If one over 56 is the first term
and negative one over 78 is the second term, there is some 𝑟-value that we multiply
the first term by to get the second term. And the same thing is true for the
third term. We could take the second term,
multiply it by the same 𝑟-value, and get the third term. This value when we’re working with
geometric sequences is called a common ratio. And we use the formula 𝑎 sub 𝑛
equals 𝑎𝑟 sub one times 𝑟 to the 𝑛 minus one power.
This means if we know the first
term and the common ratio, we can find any term in the sequence. But since we know the second and
first term, we can use this formula to help us find the common ratio. The second term is equal to the
first term multiplied by the common ratio to the two minus one power which is just
the common ratio, the ratio to the first power. That means negative one over 78
equals one over 156 times 𝑟, which when we simplify that is 𝑟 over 156. In order to solve for 𝑟 here, we
wanna multiply both sides of the equation by 156.
On the right, that cancels out
leaving us with 𝑟. And negative 156 over 78 equals
negative two. The common ratio for this sequence
is negative two. Using this information, we can now
solve what term number 16 over 39 would be equal to. We know, for this geometric
sequence, 𝑎 to the 𝑛 term is equal to the first term times negative two to the 𝑛
minus one power. If we plug in 16 over 39 for 𝑎 sub
𝑛 and one over 156 for the first term to solve the problem, we’ll then need to
isolate this 𝑛.
If we multiply one times negative
two to the 𝑛 minus one power, we can rewrite our problem like this. And again, we’ll multiply both
sides of the equation by 156. When we do that, we get 64 is equal
to negative two to the 𝑛 minus one power, which at first glance seems very
difficult to solve. You might think we could solve this
with a log, but we can’t do that because we’re working with a base that’s
negative. But we are on the right track
there. Can we rewrite 64 with a base of
negative two to some exponent?
We know that we’ll only be dealing
with even-numbered exponents. Otherwise, it would not produce the
even number 64. So what about negative two to the
fourth power? That’s only equal to 16. So we go up by two to keep our
exponent even. Negative two to the sixth power
does in fact equal 64. And since that is the case, we can
rewrite 64 as negative two to the sixth power. Once both sides of our equations
have the same base, we can solve for their exponents by setting their exponents
equal to one another. Six equals 𝑛 minus one.
And that means 𝑛 equals seven. What does this seven mean
exactly? It means that 16 over 39 is 𝑎 sub
seven. It’s the seventh term in this
geometric sequence. Since our question is only asking
the order of the term, the final answer is seven. It’s the seventh term.
