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## Algebra 1

### Course: Algebra 1 > Unit 9

Lesson 3: Introduction to geometric sequences- Intro to geometric sequences
- Extending geometric sequences
- Extend geometric sequences
- Extend geometric sequences: negatives & fractions
- Using explicit formulas of geometric sequences
- Using recursive formulas of geometric sequences
- Use geometric sequence formulas

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# Using recursive formulas of geometric sequences

Sal finds the 4th term in the sequence whose recursive formula is a(1)=-⅛, a(i)=2a(i-1).

## Want to join the conversation?

- I wonder what the logic is at2:19; how did Sal generalize into a(i)=(-1/8)(2^(i-1))?(56 votes)
- Another way to think of it is that every time you need a new term, you multiply by 2. If you have an original number of 3, your term numbers
`i`

would look like this top row. The row below would be your values:`1 2 3 4 5 6`

`3 6 12 24 48 96`

You can write a quick, general formula from this for all geometric sequences:`first value`

x`multiplier`

raised to`number of the term, minus one`

So it is`a(i) = a(1) ∙ (2) ^ (i - 1)`

If you don't adjust the exponent by one, you will find terms that are in the wrong location.

This is sometimes called the explicit formula, because you can generate**any**term if you know the first value and multiplier (common ratio).

In this simplified case I showed above, a(1) is 3

If we want to find the 4th term, here is how we calculate it:

a(4) = 3 ∙ (2) ^ (4 - 1)

or a(4) = 3 ∙ (2) ^ 3

or`a(4) = 3 ∙ (2)³`

That means the 4th term is 3 ∙ 8 or 24

Bingo!

So for Sal's example, the terms are messier and we start out knowing**only**the first value and the multiplier, and the important information that it follows the rules for a geometric sequence.`1 2 3 4 5 6`

`- 1/8 ? ? ? ? ?`

Each term is 2 times the previous. So I can reuse most of my equation from my simple example:`a(i) = a(1) ∙ (2) ^ (i - 1)`

For the fourth term

a(4) = - ⅛ ∙ (2) ^ (4 - 1)

a(4) = - ⅛ ∙ (2) ^ (4 - 1)`a(4) = - ⅛∙ (2)³`

which is a(4) = - ⅛∙ 8

That ends up with`-1`

for the 4th term. Hope that helps(72 votes)

- Why is [Sal] Using "i" for the "current value" instead of using "n"?(14 votes)
- The most common variables used for indices are i, j, k, m, n. It does not always have to be n. In this case it is i. You will see other examples here on Khan that use j, k and m as well.(16 votes)

- He is putting the power inside the parentheses again, is he doing it right?(11 votes)
- Since 2 is positive, it does not matter if it is inside or outside

It would be better if he stayed consistent, but they are the same

the main reason you put it on the outside is if the number inside is negative which does make a difference

(-2)^2 is not the same as -2^2

Hope this helps(13 votes)

- Shouldn't you put the exponent always outside of the parenthesis when solving?(7 votes)
- In the case above, yes. But more generically, it depends on what you're multiplying. Note that

is not equal to`(-4)^2 = -4 * -4 => 16`

`(-4^2) = -1 * (4*4) => -16`

(7 votes)

- Is there 1 specific formula you use in geometric sequences?(5 votes)
- I know I'm a little late, but the formula is actually aₙ = a₁ * rⁿ ⁻ ¹.(1 vote)

- So if a geometric series is the sum of the terms of the sequences, then Arithmetic series are the same. [correct me if am wrong](4 votes)
- An arithmetic sequence is a sequence with the difference between two consecutive terms constant. The difference is called the common difference. Recall a difference is subtraction so (x2-x1).

Example: 1, 5, 9, 13, 17, ... where the common difference is 4

A geometric sequence is a sequence with the ratio between two consecutive terms constant. This ratio is called the common ratio. Recall a ratio is x2/x1.

Example: 1, 2, 4, 8, 16, 32, ...

Where the common ratio is 2.(4 votes)

- lol he said pause the video to find it out but that is why I am watching it XD I don't know how(4 votes)
- isnt there any faster way to do it instead of just solving each expression to find a(n)(3 votes)
- For a geometric sequence with recurrence of the form a(n)=ra(n-1) where r is constant, each term is r times the previous term.

This implies that to get from the first term to the nth term, we need to multiply by n-1 factors of r.

Therefore, for a geometric sequence, we can calculate a(n) explicitly by using a(n)=r^(n-1)*a(1).(1 vote)

- After watching a lot of these videos, it seems like sequences are functions. Is this true?(2 votes)
- Yes this is because a sequence has 1st, 2nd, 3rd, etc. terms, so it will not repeat.(3 votes)

- I don't understand the logic of how the expressions are written, it's very very confusing. Specifically, saying that a(i) = 2a(i)-1

I understand this might be the way it's done, but isn't there some better way to express the idea? This way isn't logically consistent. (meaning that if we substitute any value for a(i) and then look at the 'equation', it is false. I've been having a heck of a time getting around that when trying to solve these.(3 votes)- It was not saying that a(i)=2a(i)-1.

What it said was a(i)=2a(i-1). So if i=5, it says a(5)=2a(4). In general, it's saying that each term is double the term before it, since a(i-1) is always the term before a(i).(3 votes)

## Video transcript

- [Voiceover] The
geometric sequence a sub i, is defined by the formula
where the first term, a sub one is equal to negative 1/8, and then every term after
that is defined as being, so a sub i is going to be two
times the term before that. A sub i is two times a sub i, minus one. What is a sub four, the
fourth term in the sequence? Pause the video, and see
if you can work this out. Well there's a couple of ways
that you could tackle this. One is to just directly
use these formulas. We could say that a sub four, well that's going to be
this case right over here. A sub four is going to be
equal to two times a sub three. Well, a sub three, if we
go and use this formula, is going to be equal
to two times a sub two. Each term is equal to two
times the term before it. Then we go back to this formula again, and say a sub two is going
to be two times a sub one. Two times a sub one. Lucky for us, we know that
a sub one is negative 1/8. It's going to be two times negative 1/8, which is equal to negative 1/4. Negative 1/4. So this is negative 1/4. Two times negative 1/4,
is equal to negative 2/4, or negative 1/2. A sub four is two times a sub three. A sub three is negative 1/2. So this is going to be
two times negative 1/2, which is going to be
equal to negative one. So that's one way to solve it. Another way to think about it is, look, we have our initial term. We also know our common ratio. We know each successive term is two times the term before it. So we could explicitly, this is a recursive definition
for our geometric series. We could explicitly write it as a sub i is going to be equal to our
initial term negative 1/8. Then we're going to multiply it by two. We're going to multiply it
by two, i minus one times. So we could say times two,
to the i minus oneth power. Let's make sure that makes sense. A sub one, based on this formula, a sub one would be negative 1/8, times two to the one minus one. Two to the zeroeth power. So that makes sense. That would be negative 1/8. Based on this formula, a sub two would be negative 1/8, times two to the two minus one. So two to the first power. We're going to take our initial term, and multiply it by two, once. Which is exactly right. A sub two is negative 1/4. So we want to find the
fourth term in the sequence, we could just say well,
using this explicit formula, we could say a sub four,
is equal to negative 1/8, times two to the four, minus one. Two to the four, minus one power. So this is equal to negative 1/8, times two to the third power. This is negative 1/8 times eight. Negative 1/8 times eight,
which is equal to negative one. You might be a little bit,
a toss up on which method you want to use, but for
sure this second method, right over here where we'd come
up with an explicit formula once we know the initial term, and we know the common ratio, this would be way easier, if
you were trying to find say, the 40th term. Because doing the 40th term
recurs to be like this, would take a lot of time,
and frankly, a lot of paper.