>> Thank you for joining us for the Cypress College Math Review
on arithmetic and geometric sequences.
Objective one, sequences in general.
A sequence is an ordered collection
of numbers or objects.
The notation that's commonly used is a sub 1 is the first
term, a sub 2 is the second term, a sub 3 is the third term,
and so on and so forth.
[variable] a sub n is the nth term or the general term.
Example, describe the sequence and determine the next object.
We have a triangle and a square and a pentagon and a hexagon.
Let's see.
Well, this is a sequence of what are called polygons.
The first object has three sides.
It's a polygon with three sides.
Then we have one with four sides
and then five sides and then six sides.
So, the next object
in the sequence should be a polygon with seven sides.
So, this is a sequence of polygons,
and each polygon has one more side.
In this example, we're asked to describe the sequence
and determine the next term.
We have 1, 4, 9, 16, 25.
I find it's often easier to determine the pattern by looking
at the later numbers
in the sequence rather than the first few.
So, we look at 9, 16, 25.
Hmm, well, 16 is 4 squared, and 25 is 5 squared.
Does that work for the other numbers?
Well, 9 is 3 squared.
And 4 is 2 squared and 1 is 1 squared.
Ah, so these are all perfect squares.
So, that's describing the sequence.
They're all perfect squares.
Now, what would the next term be?
So, we have 1 squared, 2 squared,
3 squared, 4 squared, 5 squared.
So, the next number would be 6 squared, which is 36.
So, the next term is 36.
There's another thing that we can do for this pattern.
We can actually determine a formula
for the terms in this sequence.
Let's use the notation we described earlier.
So, a sub 1 is 1 squared,
a sub 2 with the subscript 2 is 2 squared.
[variable] a sub 3 is 3 squared.
[variable] a sub 4 is 4 squared.
[variable] a sub 5 is 5 squared.
So, this number, the subscript matches the base.
So, if we have a sub n for the general term in the sequence,
that would be the base of this exponential expression.
So, a sub n would be n squared.
This is the formula
that represents the nth term in the sequence.
So, a sub n is equal to n squared.
That's a formula that represents the terms in the sequence.
Let's say we want to figure out then the 10th term.
So, we use this formula.
And now we want a sub 10, the 10th term in the sequence.
So, a sub 10 would be 10 squared, and 10 squared is 100.
The 10th term of the sequence is 100.
This sequence has the number 1, 1, 2, 3, 5, 8, 13, 21, 34.
What's the pattern here?
Hmm, well, we have 3 plus 5 is 8.
Five plus 8 is 13.
Eight plus 13 is 21.
That works, doesn't it?
Let's check it out from the beginning.
If we take these two numbers here, the first two terms
in the sequence and we add them up,
we get the third number, don't we?
1 plus 1 is 2.
Let's try and see if that pattern works for all of them.
1 plus 2 is 3.
2 plus 3 is five.
3 plus 5 is 8.
5 plus 8 is 13.
Yes, that pattern works.
This is called the Fibonacci sequence.
It's named after the guy that came up with it.
It's a very special sequence.
It actually occurs a lot in the study of botany,
computer science, genetics and many other fields.
Let's find the 11th term of this sequence.
Let's list out the terms.
So, the first term, a sub 1 is 1.
[variable] a sub 2 is also 1.
[variable] a sub 3 is 2, a sub 4 is 3
because it's the fourth term.
[variable] a sub 5 is 5.
[variable] a sub 6 is 8.
[variable] a sub 7 is 13.
[variable] a sub 8 is 21.
[variable] a sub 9 is 34.
Now, how do we get a sub 10?
Well, to get a sub 10, I would have
to add the previous two terms of the sequence.
21 plus 34 is 55.
So, a sub 10 is 55.
[variable] a sub 11, I need to add 34 and 55 and we get 89.
The 11th term of the Fibonacci sequence is 89.
In this problem, we're given the formula, a sub n equals n
over the quantity n plus two.
And we're supposed to find the first four terms.
What does that mean, find the first four terms?
It means to find a sub 1, which means plug in 1 for n,
and then find a sub 2, and so on and so forth.
So, here's our formula.
And we first want to find a sub 1.
We plug in 1 for n. And we have 1 over the quantity 1 plus 2.
Can you cancel the ones?
No. You need to do what's in the denominator first.
1 plus 2 is 3.
So, we get one third.
That does not reduce.
So, a sub 1 is one third.
Now, we want to find the second term, the second term
of the sequence, we plug in 2 for n. We have 2
over the quantity 2 plus 2.
We do what's in the denominator first
because of our order of operations.
Order of operation says, do what's in the numerator,
do what's in the denominator and then reduce if possible.
So, we have to add the 2 plus 2, and we get 4.
Now, can we reduce two 4s?
Yes. Two 4s reduce to one half.
[variable] a sub 2 is one half.
[variable] a sub 3, we substitute in 3 for n
in the denominator, we get 3 plus 2, which is 5.
And three fifths does not reduce.
[variable] a sub 3 is three fifths.
[variable] a sub 4, we substitute in 4 for n.
In the denominator, we get 4 plus 2, which is 6.
And we have 4 over 6.
Four over 6 does reduce by dividing the numerator
and denominator by 2, and we get two thirds,
a sub 4 is two thirds.
Therefore the first four terms are one third, one half,
three fifths and two thirds.
Pause the video and try these problems.
[ Pause ]
Objective two, successive differences.
We can use what are called successive differences
to find the pattern for some sequences.
In order to do this, we find the difference between each pair
of consecutive terms in the sequence.
Then we find the difference between each pair
of the original differences.
We continue this process
until we discover a pattern in the differences.
We can use this pattern to determine the next term.
Let me show an example.
Use successive differences
to determine the next term in the sequence.
Our sequences 3, 5, 8, 12, 17.
Here's our sequence.
So, difference means to subtract.
So, 5 minus 3 is 2.
8 minus 5 is 3.
12 minus 8 is 4.
17 minus 12 is 5.
So, we found the difference
between each successive terms in the sequence.
Now, we're going to do this again.
Find the difference between those differences.
3 minus 2 is 1.
4 minus 3 is 1.
5 minus 4 is 1.
Bingo, we have a pattern.
Okay, so now we have a pattern.
We know it's going to be a 1.
All right, so we know that the next number
down the bottom is going to be a 1.
So, how do we figure out the next number here?
Well, let's see.
5 minus 4 is 1.
So, you would subtract to get the number down here.
So, what do you do to figure out what this number is here?
You would add, wouldn't you?
Yes. 4 plus 1 is five.
3 plus 1 is 4, exactly.
So, we're going to add to figure
out what number goes right there, aren't we?
Yes, we're going to add.
So, 5 plus 1 is 6.
Now, we do the same thing for the number that's missing
up here, which is our answer to our problem.
17 plus 6 is 23.
The next term in our sequence is 23.
Use excessive differences
to determine the next term in the sequence.
And our sequence here is 2, 9, 28, 65, 126.
So, we subtract, 9 minus 2 is 7.
28 minus 9 is 19.
65 minus 28 is 37.
126 minus 65 is 61.
And we don't see a pattern yet.
At least, I don't.
All right, next line.
19 minus 7 is 12.
37 minus 19 is 18.
61 minus 37 is 24.
Ah, those all different by 6.
Yeah, 18 minus 12 is 6, and 24 minus 18 is 6.
All right.
Let's put up the next grid, we know that the next --
on the next line, we will have a 6.
So, we add 24 plus 6 and get 30.
Thirty goes here.
We add 61 plus 30 to get 91 here.
And then to get our answer, which is the next term
of our sequence, we had 126 plus 91 to get 217.
The next term of our sequence is 217.
Pause the video and try these problems.
[ Pause ]
Objective three, arithmetic sequences.
In this example, we're going to again,
use successive differences
to determine the next term in that sequence.
Sequence is 9, 14, 19, 24, 29.
We subtract 14 minus 9 is 5.
19 minus 14 is 5.
24 minus 19 is 5.
29 minus 24 is 5.
And here we quickly determine a pattern.
We add 5 to each term to get the next term.
So, the next term in our sequence is 34.
Any sequence where there is a common difference
between successive terms, that's called an arithmetic sequence.
We're going to use d to denote the common difference.
So, in the above example, d is five.
Let's determine a formula for this sequence.
So, the first term, a sub 1 is 9.
[variable] a sub 2 is 14.
[variable] a sub 3 is 19.
[variable] a sub 4 is 24.
And, a sub 5 is 29.
How did we get to a sub 2?
Well, we took the first term,
which was 9, and we added 5 to it.
To get the third term, we added another 5.
To get the fourth term, we added another 5.
To get to the fifth term, we added another 5.
When you're looking for patterns,
don't look at the beginning, look at the end.
So, this fifth term here, we have a 9, and we have four 5s.
The fourth term, we have the 9 plus three 5s.
The third term, we have the 9 plus two 5s.
Now, we want this to be a pattern.
So, we want to make the first and the second terms
to be the same pattern.
So, even though it kind of looks dumb to write it,
we're going to write it anyhow.
So, this is 9 plus one 5s.
And this is 9 plus no 5s.
So, 9 plus zero 5s.
Good. So, we've written down the pattern that we have here.
So, we have 9 plus zero 5s.
9 plus one 5s and so on and so forth.
Now, we're going to write that into a formula.
So, we want a formula for a sub n,
the general term in the sequence.
What would we have?
[variable] a sub n would be 9.
Of course, there's always 9 plus something times 5.
Well, let's see, what have we got here?
When this is a 1, this is a zero.
When this is a 2, this is a 1.
When this is a 3, when the subscript is a 3,
the number over here is a 2.
It's always one smaller, isn't it?
Yeah. The number of 5s is always one smaller than the subscript.
So, when the subscript is a 5, then the number of 5s is a four.
When the subscript is an n, then the number of 5s is n minus 1.
So, the formula for the sequence is a sub n equals 9 plus the
quantity n minus 1 times 5.
Nine was our first term, a sub 1.
Five was our common difference.
And the common difference is normally denoted, d. So,
if we replace 9, which is a sub 1, and 5 with d,
we get the formula in general for an arithmetic sequence.
[variable] a sub n equals a sub 1 plus the quantity n minus 1
times d. This formula can be used to find terms
in an arithmetic sequence.
Example, write a formula for the nth term of the sequence
and then use it to find the 10th term.
Our sequence is negative 7, negative 4, negative 1, 2, 5,
and so on and so forth.
Negative 4 minus negative 7,
that would be negative 4 plus 7, which gives us 3.
So, the difference is three.
Negative 1 minus negative 4 would be negative 1 plus 4.
And that difference is also 3.
So, we right now know that this is an arithmetic sequence
where the common difference is 3.
Let's go ahead and do it for the other 2 terms.
But we already know that's arithmetic
since we've checked it twice.
2 minus negative 1 is also 3.
And 5 minus 2 is three.
You don't have to do the successive differences
if you've already checked it twice,
you know that it's an arithmetic sequence.
All right.
So, we know it's an arithmetic sequence.
We know the common difference is three.
So, we can apply this formula.
[variable] a sub 1 is negative 7, and d is 3.
What are we looking for?
A formula?
We're looking for a formula for the nth term, all right?
So, we start off with this formula.
[variable] a sub 1 is negative 7, and d is three.
So, we substitute in negative 7 for a sub one.
We substitute in 3 for d. And we should simplify our formula.
We distribute the 3.
So, we get negative 7 plus 3n minus 3.
So, a sub n is equal to 3n minus 10.
And that's the formula we were looking for,
which is the first answer for our problem.
The second question was use it to find the 10th term.
So, we're looking for a sub 10.
[variable] a sub 10 would be 3 times 10 minus 10, which is 20.
The 10th term of the sequence is 20.
Write a formula for the nth term of the sequence
and use it to find the 12 term.
The sequence is 20, 16, 12, 8, 4.
We write it out.
16 minus 20 is negative 4.
12 minus 16 is negative 4.
We know we have an arithmetic sequence.
And d is negative 4.
There's no reason to do the rest of the successive differences.
We know we have an arithmetic sequence and d is negative four.
Here's the formula that we use.
We know that a sub 1 is 20, and d is negative 4.
Let's substitute those in.
So, a sub n is equal to 20 plus the quantity n minus 1 times --
now, be careful to put d in parentheses
because d is negative.
Now, we're going to distribute the negative 4.
So, we get 20 minus 4n plus 4.
So, we get a sub n is equal to 24 minus 4n.
This is the formula that we were asked for in the problem.
So the formula for the nth term
of the sequence is a sub n equals 24 minus 4n.
That's the first answer that we were asked for in the problem.
Next, we were asked for the 12th term.
To find the 12th term, we would substitute in n12
for n. [variable] a sub 12 would be 24 minus 4 times 12.
We'd put in 12 for n, and we get negative 24.
The 12th term of the sequence is negative 24.
Determine the number of terms in this sequence using the formula.
7, 10, 13 and so on and so forth to 46.
Well, 10 minus 7 is 3.
So, the common difference is 3, right?
But, wait a minute, you have to check it twice,
you always have to check it twice.
13 minus 10 is also 3.
Okay, so we have an arithmetic sequence, and d is three,
the common difference is 3.
So, we can use this formula, a sub 1 is 7.
And d is 3.
Now, we're used to finding a particular term in the sequence.
We're not doing that this time.
The term is 46.
We're trying to figure out which term it is.
Now, we could sit there and go at 3, at 3, at 3, at 3,
at 3 and figure out which term it is.
But what if instead of 46, it was 4,600 blah, blah, blah?
No, no, no, no, no.
We're supposed to show the math,
we're supposed to use the formula.
We're not going to get any credit
if we don't use the formula.
All right, so here's the formula.
And the nth term is 46.
The goal is to figure out what n is.
So, the nth term is 46.
[variable] a sub one, the first term is 7,
and the common difference is 3.
Our goal is to solve this equation
for n. Let's distribute the 3.
So, we get 3n minus 3 there.
So, on the right hand side, we have 3n plus 4.
We subtract 4 from both sides.
So, we have 42 equals 3n.
Then we divide both sides by 3, and we get 14 equals n. So,
46 is the 14th term of the sequence.
[variable] a sub 14 is equal to 46.
So, this sequence has 14 terms.
Determine the number of terms in the sequence using the formula
and the sequence is 8, 3, negative 2,
so on and so forth, to negative 77.
So, we subtract 3 minus 8 is negative 5.
We have to check it twice.
Negative 2 minus 3 is negative 5.
So, we have an arithmetic sequence.
And the common difference, d, is negative 5.
Here's our formula.
[variable] a sub 1 is eight, and d is negative five.
The last term of our sequence is negative 77.
We want to find out how many terms are in this sequence.
So, we want to find out which number term is negative 77.
So, a sub what is equal to negative 77?
So, we substitute in negative 77 for a sub n. And we want
to find out what n is.
How many terms are in the sequence,
which term number is it?
So, a sub 1 is 8, and d is negative 5.
Be sure to put negative 5 in parenthesis since it's negative.
On the next line, we distribute the negative 5.
So, we get negative 5n plus five.
Now, 8 plus 5 gives us 13.
So, we have negative 77 equals 13 minus 5n.
We subtract 13 from both sides
to get negative 90 equals negative 5n.
Divide both sides by negative 5, and we get 18 equals n. So,
negative 77 is the 18th term of this sequence.
So, the sequence has 18 terms.
The first row in a stadium has 20 seats.
The second row has 24 seats.
The third row has 28 seats.
This pattern continues to the back of the stadium.
The stadium has 52 rows, how many seats are in the 25th row?
So, the sequence is 20, 24, 28, and so on and so forth.
[variable] a sub 1, the first term in the sequence is 20.
[variable] a sub 2 is 24.
[variable] a sub 3 is 28.
What are we looking for?
We're looking for how many seats are in the 25th row,
which would be, a sub 25.
Is this a particular type of sequence?
Yes, it is arithmetic.
24 minus 20 is 4.
Also, 28 minus 24 is also 4.
This is an arithmetic sequence
that has a common difference of 4.
So, a sub one is 20 and d is 4.
Here's our formula.
And we're looking for a sub 25.
So, we substitute in 20 for a sub 1.
We substitute in 4 for d. And we also substitute in 25 for n,
since we're looking for the 25th term in the sequence.
We do what's in the parenthesis first.
25 minus 1 is 24.
We multiply next, 24 times 4,
and then we add the 20, to get 116.
We were looking for a sub 25 and we got 116.
There are 116 seats in the 25th row of the stadium.
Pause the video and try these problems.
[ Pause ]
Objective four, arithmetic series.
A series is the sum of the terms of a sequence.
So, for a series you add up the terms of a sequence.
The German mathematician, Carl Gauss at the age of 10,
was asked to add up the first 100 natural numbers.
Now, what's a natural number?
Those are the counting numbers; 1, 2, 3,
4 and so on and so forth.
So, he was asked to add up these numbers here.
He noticed the following pattern.
If you take the first number 1, and the last number 100
and add them up, you get 101.
And if you take the next number, 2 and the next
to last number, 99 you also get 101.
In fact, if you take 3 and 98, you also get one 101.
And if you take 4 and 97, you also get 101.
And if you take 5 and 96, you also get 101.
And in fact, every single pairing like that, you're going
to get 101 every single time.
How many pairs are like that?
Every single pair is going to be like that.
In fact, how many pairs are there?
There's going to be 50 of those pairs,
since there's 100 numbers, aren't there?
There's going to be 50 of those pairs,
and every single pair is going to add up to 101.
So, what's the sum?
The sum is 50 times 101 which turns out to be 5,050.
Kind of a cool thing for a 10 year old to do.
We now want to find a formula for adding
up a certain number of natural numbers.
So, we are going to add up 1 plus 2 plus 3, plus 4,
and so on and so forth.
The last number is just going to be n, it's not going
to be 100 it's going to be just some specific natural number.
We don't know what it is.
Now, what would the number right before n be?
Well, since it's a natural number,
the number right before it would be 1 smaller than it.
So, that the n minus 1.
And the number right before that would be n minus 2, yeah?
Okay. Now, we're going to take that whole sum, and we're going
to write it backwards.
So, instead of from the left, going 1, 2, 3.
Now, from the right, we're going to go 1 plus 2 plus 3
and so on and so forth.
And then we're going to have the n minus 2, and then we're going
to have the n minus 1, and then we're going to have the n.
And then the left hand side, we still have some equals.
Now, we're going to add up these two sums.
So, on the left hand side,
we have sum plus sum equals twice the sum, okay?
So, we're going to add up these two equations.
Now, what do we get when we add up these?
1 plus n is just n plus 1.
We can write it in any order.
1 plus n or n plus 1 doesn't make a difference.
What do we get when we add 2 plus n minus 1?
Well, 2 minus 1 is 1.
So, I would get n plus 1.
Oh, that's the same thing, cool.
What do you get when you add 3 and n minus 2?
Well, 3 minus 2 is positive 1.
So, I get n plus 1.
Huh, the parentheses are all the same.
Then, let's come down here.
[variable] n minus 2 and 3, I get the same thing; n plus 1.
And n minus 1 and 2, when I add those up, I get n plus1.
[variable] n plus 1 is n plus 1.
Wow, I get the same thing
in every single one of the parentheses.
So, when I add up these two things on the right hand side,
every single time I get n plus 1.
And I have the little dot, dot, dot in the middle showing
that I get the same thing every single time all the way
across the board.
Now, how many sets of parentheses are there?
You're going to say 6, right?
No, no, no, no, no, there's a little dot dot dot in the middle
that says there's a whole bunch of [inaudible] in between
that we're not seeing, right?
We can't write them all.
How many of them are there?
Well, let's look them up top; 1, 2, 3.
And then so on and so forth to n. There were n numbers
across here, weren't there?
Yes, there were n natural numbers
that we're trying to add up.
So, there would be n sets of parentheses down here.
So, on the right hand side, we have n times
that parenthesis, n plus 1.
So, our equation is twice the sum is equal
to n times the parenthesis n plus 1.
If we divide both sides by two, we have our formula.
The sum of the first n natural numbers is equal to,
n times the quantity n plus one all over 2.
And there's our form of that we're looking for.
Now, since we have Carl Gauss's example,
let's go ahead and check it out.
He was adding up the first 100 natural numbers.
So, let's put in 100 for n. So,
we have 100 times the quantity, 100 plus 1 over 2.
So, that's 100 times 101 all over 2.
And I don't need to, but I could take that 100 over 2,
and pull that over to the side.
100 divided by 2 is 50.
Yeah, there were 50 of those 101 pairings, weren't there?
And that gives me the exact same answer 5,050.
Determine the sum of the first 500 natural numbers.
And we are, of course, supposed to use a formula here.
So, here's the formula that we just developed.
So, n is equal to 500.
So, we have 500 times the quantity 500 plus 1 all over 2.
So, we have 500 times 501, all divided by 2.
And we get 125,250.
The sum of the first 500 natural numbers is 125,250.
Determine the following sum 42 plus 43 plus 44,
and so on and so forth to 95.
Well, this doesn't match the pattern we had before.
Because before, we had 1 plus 2, plus 3.
So, we have to make it match that pattern to be able to work
with this, because that's the only formula we have.
So, we need 1 plus 2 plus 3.
So, 1 -- let's see.
Well, 42 is 41 plus 1.
And 43 is 41 plus 2.
Yeah, I want to make this look like 1 plus 2 plus 3.
So, if I wrote 42 as 41 plus.
And 43 as 41 plus 2.
And 44 would be 41 plus 3, so on and so forth.
Now, I simply subtract 41 from 95 and get 54.
So, I have it written out.
Now, what am I going to do?
I'm going to take out all those 41s.
How many 41s are there?
One, two, three, and so on and so forth.
There are 54 of those, aren't there?
There are 54 41s.
So, I'm going to take out 54 41s.
So, I'm going to have 54 times 41 plus the sum, 1 plus 2,
plus 3 and so on and so forth to 54.
Oh, we know how to add that up because we have a formula.
Here's our formula.
And we're going to use that to add up the numbers
that are in the parenthesis.
[variable] n is 54.
So, we multiply 54 times 41 to get 2,214.
We substitute in 54 for n. So,
we have 54 times the quantity 54 plus 1
on top and divide it by 2.
So, we have 55 in the parenthesis, the numerator.
We get 2,970.
We divided to get 1,485.
And finally, we add to get 3,699.
Our sum is 3699.
Pause the video and try these problems.
Objective five, geometric sequences.
Consider the sequence 3, 6, 12, 24, 48, and so on and so forth.
Well, 6 minus 3 is 3.
So, it's an arithmetic sequence where d is 3, right?
Wait a minute, we've got to check it twice, don't we?
12 minus 6 is 6.
So, this is not arithmetic.
[variable] d is not 3.
That doesn't work.
And in fact, the pattern here is that we multiply each term by 2
to get to the next term.
Another way to put that is if you divide any term
by the previous term, we get the same result.
So, let's take 6 and divide it by the term that comes before 3.
And that gives you 2.
Let's take 12 and divide it by the term
that comes before it, 6.
And we get 2.
Ah, we checked it twice, so we're good.
So, this is a geometric sequence.
And the common ratio, which we call 4 r is 2.
Let's check it a third time just for fun.
24 divided by the term right before it, 12 is 2, again.
A sequence that has a common ratio is called a geometric
sequence, and we call the common ratio, r. So,
this is a geometric sequence.
And the common ratio for this sequence is 2.
Determine the next term in the sequence.
And our sequence is 5, 15, 45 and so on and so forth.
We notice that if we take 15 and divide it
by the term before it, which is 5, we get 3.
Also, if we take 45 and divide it by 15, we also get 3.
Therefore, this is a geometric sequence
with a common ratio r of 3.
To find the next term then, we would multiply the last term,
45, by the common ratio 3.
45 times 3 is 135.
The next term of the sequence will be 135
because you multiply each term by three
to get the next term in the sequence.
Write a formula for the nth term of the sequence
and use it to find the 7th term.
Our sequence is 5, 10, 20, 40 and so on and so forth.
If we take 10 and divide it by 5, we get 2.
Likewise, if we take 20 and divide it by 10, we also get 2.
We checked it twice.
So, we know that this is a geometric sequence,
and the common ratio is 2.
We were asked this time for a formula for the nth term.
So, we need to look for a pattern.
The first term, a sub 1 is five.
The second term, a sub 2 is 10.
And how do we get 10?
We take the initial term, 5, and multiply it by 2.
To get the third term of 20, we take
and multiply the 10 by another 2.
So, we have 5 times 2, times another 2.
To get the 40, we take the initial 5 and multiply it
by 2 and 2 and 2 again.
Recall that 3 to the fourth power means three times itself
4 times.
[variable] r to the fourth then means r times r times r times r.
We're looking for a pattern.
Forty is 5 times 2 times 2 times 2.
So, we can rewrite that as 5 times 2 to the third power.
I'll leave it on the outside of the parenthesis
because sometimes we'll end up with a number
in the parentheses that's negative,
and then you would need it outside the parenthesis,
the exponent.
So, we have 5 times 2 to the third power for a sub 4.
[variable] a sub 3, the third term then,
would be 5 times 2 to the second power.
We have 2 times 2, or 2 to the second power.
We want to make them all look the same.
So, the second term could be written
as 5 times 2 to the first power.
And recall that anything
to the zero power is 1 other than zero itself.
So, I know it looks silly, but we could write 5
as 5 times 2 to the zero power.
We're trying to make it so that we can see a pattern.
So, we have 5 times 2 to varying powers.
What would a sub 8 be?
Well, the exponent is one smaller than the subscript.
So, that would be 5 times 2 to the 7th power.
[variable] a sub 30 then would be 5 times 2
to the 29th power, one smaller.
So, here we go with our formula, a sub n, the general term.
The general term would be 5 times 2 to the power
of one smaller, n minus 1.
So, the formula for the nth term of the sequence,
a sub n is equal to 5 times 2 to the power of n minus 1.
Now, let's find the seventh term in the sequence.
We have our formula, we're looking for a sub 7.
So, we substitute 7 in for n. So, we have 5 times 2
to the power of 7 minus 1.
And what do we do first in our order of operations?
We do the exponent first.
So, 7 minus 1 is 6.
Now, we take our calculator.
Here's a calculator.
Now, there's different ways you can do this.
You could take 5 times --
your calculator knows the order of operations.
Your calculator knows to do the exponents first.
So, you could put in the base of 2.
And this is the exponent key, 2 to the power
of -- 2 to the power of 6.
So, it knows to do exponents before multiplication,
and then press Enter.
We have an answer of 320.
If you're not comfortable with this,
let me show you another way to do the problem.
Do the exponents first.
You know the order of operations,
you need to do the experiments first.
So, you do the base 2 to the power of --
now, put in your exponent of 6, okay?
We press Enter, and we have an answer of 64.
So, 2 to the 6 is 64.
Now, we're supposed to multiply that by 5.
So, times 5.
The calculator -- remember that I had 2
to the 6 in the calculator.
Now, we press Enter and we have 320.
So, we have 2 to the 6 is 64 and then we multiply that by 5
and we still get the answer of 320.
The 7th term of the sequence is 320.
Let's use the example we just finished to come
up with a general formula for terms of a geometric sequence.
We had a sub n equals 5 times 2 to the power of n minus 1.
And in our sequence, a sub 1 was 5, and r was 2.
Well, since a sub one is 5, we can replace 5 with a sub 1.
And we can replace 2 with r. So,
in a geometric sequence terms can be found using the formula a
sub n equals a sub 1 times r to the n minus 1 power.
Find a formula for the nth term of the following sequence.
Use the formula to find the 9th term.
Our sequence is 80, 40, 20, 10 and so on and so forth.
So, we take 40 and divide it by the term
that comes before it, which is 80.
We reduce the fraction 40 over 80
to the lowest terms and we get one half.
Now, we take the term 20 and divide it by the term
that comes before it, which is 40.
20 divided by 40 reduces also to one half.
Since they both reduced to one half,
this is a geometric sequence
and the common ratio or r is one half.
[variable] a sub 1 for our geometric sequence is 80,
the first term.
[variable] r, the common ratio is one half.
Here's the formula we just came up with.
[variable] a sub n equals a sub one the first term times r,
the common ratio to the power of n minus 1.
So, for our sequence, a sub 1 is 80, and r is one half.
The formula for the nth term is a sub n equals 80 times the
quantity one half to the n minus 1 power.
In this case, you do need the one half to be in parenthesis,
because you need the whole fraction to be raised
to the n minus 1 power.
Now, we wish to find the 9th term of the sequence.
Here's our formula.
We want a sub 9.
So, we put in 9 for n, what do we do first?
We do the exponent.
So, 9 minus 1 is 8.
Now, think order of operations,
you need to do the exponents first.
So, you need to take the 8th power of one half.
Let's use a calculator to help us out with this.
So, we want to take the 8th power of one half.
So, on this calculator, it's going to look like this.
Parenthesis, 1, and then we're going
to use the fraction button.
The fraction button on this calculator looks like this.
On the TI-30XIIS, the buttons look like this.
So, I'm going to show out the buttons on the TI-30XIIS also,
because that's the one that a lot
of students will use in our classes.
So, parenthesis, 1 and then the fraction button.
So, in this calculator, it actually looks like a fraction.
And then you put in the 2 for the denominator.
Now, we need to get out of the denominator.
On our calculator, we can just close the parenthesis.
But on this calculator, we have to get out of the parenthesis,
so we have to use the right arrow key.
So, now I close my parenthesis, okay?
So, we have parenthesis, one half, close parenthesis.
Now, we need to raise that to the 8th power.
So, here's our power key.
So, 2 to the power of 8.
And then you can press Enter, and we get 1 over 256.
So, one half raised to the 8th power is 1 over 256.
We still need to multiply that times 80.
The calculator remembers our previous result, right there.
So, we don't have to reenter it.
So, now I just press multiplication.
And then we put in 80 and press Enter.
And we get 80 over 256.
But the calculator has already reduced that fraction.
It took 80 over 256 and reduced it to 5 over 16.
The 9th term of this sequence is 5/16.
Find a formula for the nth term of the following sequence.
Use the formula to find the 8th term.
And our sequences 6, negative 18, 54, negative 162,
and so on and so forth.
So, we take negative 18 and divide it
by the first term, which is 6.
That reduces to negative 3.
We take 54 and divide it by negative 18.
And we get negative 3 also.
So, this is a geometric sequence.
And the common ratio is negative 3.
So, a sub 1 is 6.
The first term is 6.
This is geometric, and the common ratio is negative 3.
The formula for a geometric sequence, the nth term is this,
a sub n equals a sub 1 times r to the n minus one.
The formula for the nth term
of this sequence is a sub equals 6 times the quantity negative 3
to the n minus 1.
The negative 3 definitely needs to be in parenthesis.
Now, we wish to find the 8th term of this sequence.
We have our formula.
We substitute in 8 for n. We look at the exponent first.
8 minus 1 is 7.
And now, we're going to need our calculator to help us out.
You can do this different ways.
You could do 6 times parenthesis --
now, do I hit this key for subtraction?
Kind of giving it away, or this key for negative?
Negative, good.
Negative, what is it?
Three. Close parenthesis.
And then to the power of 7, to the power of 7.
So, in this way, I put the whole thing
in straight from left to right.
6 times negative 3 to the 7th.
The calculator knows to do the exponent first.
It's going to take the 7th power of negative three first,
and then it's going to multiply it by 6.
So, there's our answer, negative 13,122.
Here's another way to do it, parenthesis,
negative 3, close parenthesis.
We're going to raise that to the 7th power, okay?
So, we raise that to the 7th power and we get negative 2,187.
And then we're going to take our answer and multiply it times 6.
So, times 6 equals.
Do it the way that you're most comfortable with.
Either way, the 8th term of this sequence is negative 13,122.
Pause the video and try these problems.
[ Pause ]