Geometric Sequences
This page originally created by Kevin and Ethan (2021)
A geometric sequence is a sequence where each number is a fixed multiple of the number before it. Each term is obtained by multiplying the previous term by a constant called the common ratio.
Common ratio[edit]
The common ratio is the constant of a geometric sequence. To find it, divide the second term by the first term or divide any term by the previous term.
The formula for finding it can be written as the following:
Where:
r = common ratio
a = first term
n = n^{th} term
General Form[edit]
Consider the following sequence and table:
4, 12, 36, 108
Term | Expanded | t_{n} = ar^{ n-1 } |
---|---|---|
t_{1} | First term = 4 | a |
t_{2} | (4)(3) = 12 | ar |
t_{3} | (4)(3)^{2} | ar^{2} |
t_{4} | (4)(3)^{3} | ar^{3} |
The "general term" is the apparent pattern of a sequence.
The formula for the general term is:
t_{n} = a r ^{n -1}
or
t_{n} = t_{1}r ^{n-1}
Where:
t_{n} = general term of the geometric sequence a = t_{1} is the first term of the sequence r is the common ratio n is the position of the term being considered
Common Problems[edit]
Determining if a sequence is geometric[edit]
For the sequence to be considered geometric, the difference between one term and its previous term needs to be the same.
Essentially, t_{2}/t_{1} = t_{3}/t_{2} = t_{4}/t_{3}
If the differences between each term and its previous are the same, the sequence in question is in fact geometric.
Determining the common ratio[edit]
The common ratio of a geometric sequence can be found by dividing a term by its previous term.
Given the following sequence: 6, -12, 24, -44. Find the common ratio.
Select two successive points. For this example, we chose 24 and -12.
Since r = t_{n} / t_{n-1}, we can substitute our values, 24 and -12 into t_{n} and t_{n-1}
r = 24/-12
r = -2
The common ratio is -2
Determining the formula for the general term[edit]
Once you have determined the common ratio, all you need to do is to substitute it in the formula: t_{n} = ar^{ n-1 }
Consider the sequence: 4, 12, 36, 108...
a = 4
Determine the common ratio:
r = 36 / 12 r = 3
Substitute into t_{n} = ar^{ n-1 }
t_{n} = ar^{ n-1 } t_{n} = (4)(3)^{ n-1 }
Determining the n^{th} term[edit]
To determine the n^{th} term of the geometric sequence, you can multiply the first term by the constant to the power of n.
This is written as follows:
a_{n} = ar^{n}
where r is the common ratio between terms
t_{3} = 8 and t_{7} = 128 Determine t_{10} First find the common ratio.
t_{6}/t_{2} = ar^{((7)-1)} / ar^{((3)-1)} = 128 / 8 r^{4} = 16
Fourth root both sides.
r = +/- 2
Find the first term through substitution.
t_{3} = a(2)^{((3)-1)} 8 = a(2)^{(2)} 8 = 4a a = 2
While x = 2
t_{12} = (2)(2)^{((12)-1)} t_{12} = 4096
While x = -2
t_{12} = (2)(-2)^{((12)-1)} t_{12} = -4096
Determining the number of terms in a geometric sequence[edit]
To find the number of terms in a geometric sequence, you'll need to find the first term and the common ratio between terms.
Once you have these numbers, put them in the form of t_{n} = a r ^{n-1}
Next, divide each side by the constant a
Now that the formula is in the form of t_{n} / a = r ^{n-1}, you can guess and check to solve for n. This involves substituting differing values for n until both sides are satisfied.
Alternatively, one can use Logarithms to solve for n.
This is done by pressing the log button on your calculator and solve for n by putting the right side over the left side while both are logged so they look like this:
log(t_{n} / a) / log (r)
Either of these methods will give you n which is equal to the total number of terms in the geometric sequence.
Now consider the following:
Big Bear is trying to determine the number of terms in the sequence 32, 64, 128, ... 16,384.
We can gather that 32 is the first term and 16,384 is the final so we have the values for t_{n} and a
. Also the difference between terms is a multiple of 2, that is the r value These numbers can be written in the formula as t_{n} = a r ^{n-1}
Currently, we have 16,384 = 32 r ^{n-1}
We can divide both sides by the 32 and get 512 = 2 ^{n-1}
Here are the two methods you can use a) Guess and Check We can estimate a exponent that will get us to 512 from a base of 2. 2^{10-1} = 512, Perfect! From here, we know that the number of total terms is 9. b) Logarithms We can use the logs to find the exponent a lot quicker than guessing and checking, especially when it comes to larger numbers that can get a bit messier. Once again, we need to find a exponent that will get us to 512 from a base of 2. We will use log(512) / log (32) in our calculators and will get a answer of 9. There are 9 terms in the sequence.
Solving sequence problems where the constant and initial term are unknown[edit]
Three consecutive terms of a geometric sequence are x-15, 4, x. Solve for x and predict the next term in the sequence.
Firstly, set up an equation to solve for x. We know that to find the common ratio, r, we use the formula: r = t_{n} / t_{n-1}.
Therefore, if t_{3} / t_{2} = r, t_{2} / t_{1} must also = r.
t_{3} / t_{2} = t_{2} / t_{1} x/4 = 4/(x-15)
Then, cross multiply to get:
4(4) = x(x-15)
Set one side to zero and factor.
x^{2} - 15x -16 = 0
(x- 16)(x + 1) x' = 16, -1
Now we are able to find the common ratio by substituting our values back in the sequence. We will now have two sequences since we have two roots.
When x = 16
((16) - 15), 4, (16) 1, 4, 16 r = t_{n}/t_{n-1 r = 16 / 4 r = 4 }
When x = -1
((-1)) - 15), 4, (-1) -16, 4, -1
r = t_{n}/t_{n-1 r = 4 / -16 r = -1 / 4 }
With the common ratios, all we need to do is multiply the last term with the common ratio.
When x = 16
t_{4} = (16)(4) t_{4} = 64
When x = -1
t_{4} = (-1)(-1/4) t_{4} = 1/4
Other resources[edit]
References[edit]
Geometric Sequences, Varsity Tutors,
https://www.varsitytutors.com/hotmath/hotmath_help/topics/geometric-sequences#:~:text=A%20geometric%20sequence%20is%20a,between%20consecutive%20terms%20is%20constant.&text=where%20r%20is%20the%20common,%2C1458%2C...%7D
Geometric Sequences and Sums, mathisfun,
https://www.mathsisfun.com/algebra/sequences-sums-geometric.html
Intro to geometric sequences, Khan Academy,
https://www.khanacademy.org/math/algebra/x2f8bb11595b61c86:sequences/x2f8bb11595b61c86:introduction-to-geometric-sequences/v/geometric-sequences-introduction