# Geometric Series

**This page originally created by Ameya and Farrah (2021)**

When the terms of a geometric sequence are added, the result is known as a geometric series.

ex. 2, 4, 8, 16, … is a geometric sequence

ex. 2 + 4 + 8 + 16 + … is a geometric series

The difference between a geometric series and an arithmetic series is an arithmetic series is increased by adding a constant to each term when geometric series are increased by multiplying by a constant.

ex. 24 + 12 + 6 + 3 +...+ is a geometric series

ex. 2 + 4 + 6 + 8 +...+ is an arithmetic series

## Finite Geometric Series[edit]

Before we go over how we get the basic formulas for Finite Geometric Series, we need to understand what each variable stands for. **S _{n}** is the sum of nth terms, this is the value that you calculate for if you are asked to find the total sum of n terms in a set.

**n**is the number of terms you are given,

**r**is the common ratio between each term and

**a**is the leading term of the series. Those three values are known whereas

**S**is determined. (Also do not forget to check out Infinite Geometric Series)

_{n}

If you are not given **r** but you know the first few terms, all you need to do is calculate the ratio between the first two terms. This follows the formula

It is also important to double check the ratio by testing with other consecutive terms, such as ^{t3}⁄_{t2} or ^{t4}⁄_{t3} or ^{t10}⁄_{t9} if those terms are known.

**In General:** In a geometric series of n terms

_{n}= a + ar + ar

^{2}+... +ar

^{n-1}

_{n}= ar + ar

^{2}+...+ar

^{n-1}+ ar

^{n}

You subtract the entire line of rS_{n} by S_{n} which results in all the terms cancelling except for the first and last term. This will be presented as ar^{n} minus a.

rS_{n} - S_{n} = ar^{n}-a

Then you can factor out the S_{n} from the left side leaving you S_{n} (r - 1) and factor **a** out of the right side leaving a(r^{n}- 1). Then you equal them to each other and get

^{n}- 1)

And that results in two formulas:

^{a(1-rn)}⁄

_{1-r}or Sn =

^{a(rn-1)}⁄

_{r-1}, where r ≠ 1

You can also use the formula below if you are given the last term

There are two formulas because sometimes the answer can be nicer if put into one form or the other.

If r > 1, it is a better idea to use the right formula, and if 0 < r < 1, it is better in the left formula. You can switch between the two simply by taking out the negative sign in the brackets on both the numerator and denominator

## Summation Notation[edit]

Summation Notation, also known as Sigma Notation is used to express large sums without having to list every term. Here are two standard formats of what it would look like.

The form on the left differs from the form on the right based on the index value, **k**. The index doesn't directly affect the sum but it does indicate what the first value will be when substituted into the right hand side. This also goes with the **n** value above the sigma which determines the value that you input to get the final value from the right hand side. Simply, it shows the terms of the series start at t_{1} (that you calculate using the k value) and ends at t_{n}.

If you are given a question where you need to determine the sum of nth terms where it is in the form of summation notation. Here are a few things you need to know about the variables and their purposes.

The **∑** symbol, otherwise known as sigma meaning 'sum up', is the standard way of writing a large sum. So you can think of **∑** as a replacement for **Sₙ** from the other finite geometric series formulas.

The **k** value (otherwise known as the index) determines the first term, **a** , when the **k** value is plugged into the exponent of the **r** value. Something you need to note is that the **a** and **r** values are related to the forms if the **k** value found under the sigma is directly related to the exponent k - x, where x is an integer. In the two basic examples of a geometric series shown above, when k = 0 under sigma, the exponent becomes k - 0 simplified to k. When k = 1, the exponent becomes k - 1. For example, given the formula you can determine the value of **a** if you substitute k = 1 into the k - 1 as the exponent of the r,

*Work shown:*

a = 2(-5)^{k-1}

a = 2(-5)^{1-1}

a = 2(-5)^{0}

a = 2(1)

a = 2

Now that you have found **a**, you can use this to determine the value of **r** by setting r to

r = ^{tn}⁄_{tn-1}

r = ^{2(-5)6-1}⁄_{2(-5)(6-1)-1}

r = ^{2(-5)5}⁄_{2(-5)4}

r = ^{2(-3125)}⁄_{2(625)}

r = ^{-6250}⁄_{1250}

r = -5

Now you know all the values you require to calculate the total sum of the equation. r = -5 , a = 2 and n = 6.

= ^{a(1 - (r)n)}⁄_{1-(r)}

= ^{2(1 - (-5)6)}⁄_{1-(-5)}

= ^{2(1 - (15,625))}⁄_{6}

= ^{2(-15,624)}⁄_{6}

= ^{-31,248}⁄_{6}

= -5206

S_{n}= -5206

Just like that you have found the total sum of the equation given to you!

In some cases, when the k under the sigma does NOT relate to the exponent value of r, you can not reference one of the above formulas to get the exact values.

*Ex. 1*

Compare the two equations.

Right away you can tell that the first equation can be related to this form , this means that a = 5 , r = 7 and n = 6. The difference between the first equation and the second equation is that the second equation can not be related to that basic formula. This means that not all 3 variables align with their values in the equation. In the second equation, if you plug in the 1 in place of the **k** exponent, the values you get will be a = 35, r = 7 and n = 6.

Equation 1 results in the series: 5 + 35 + 245 + 1715 + 12005 + 84035

Equation 2 results in the series: 35 + 245 + 1715 + 12005 + 84035 + 588245

What do you notice? Since the **k** value under the sigma doesn't relate to the exponent in the second equation, the 3 variables no longer fully correlate. The **r** value and **n** value of both equations remain the same but the **a** value is affected. This shows that changing the **k** value does not alter the **r** or **n** values of the equation.

How do you deal with a geometric series written in summation notation where the right side of the sigma is not in either ?

*Ex. 2*

Given , how do we find the values of **a** and **r**?

First thing you want to do is simplify the right side.

Consider 2^{3k} where the exponents are multiplied together. Separate the two exponents and you get (2^{3})^{k}. Do the same with 5^{1-2k} by separating the 1 and -2k. This gives you 5^{1} and 5^{-2k}

= (2^{3k})(5^{1-2k})

= ((2^{3})^{k})(5^{1})(5^{-2k})

Because one of the terms has a negative exponent value, apply the Integral Exponent Rule/House Rule where X^{-n} = ^{1}⁄_{Xn}

= ^{((23)k)(51)}⁄_{(52k)}

Separate the (5^{2k}) into (5^{2})^{k} and simplify.

= ^{(8k)(5)}⁄_{25k}

Now you can take the 5 from the numerator and write it besides the fraction. The fraction then becomes (^{8}⁄_{25})^{k} because the numerator and denominator share the same exponent, this is otherwise known as the Power of a Quotient Exponent Law where ^{Xm}⁄_{Ym} = (^{X}⁄_{Y})^{m}

= 5(^{8}⁄_{25})^{k}

Now you know what the right side looks like when simplified. This is what your equation looks like

As you analyze this equation you notice that the **k** value under the sigma does not align with the k - x exponent next to the **r**, this means you can not relate this equation to this form . But as we explored before, even if the **k** value does not align with the k - x exponent it does not alter the final **r** value if put in this form. It will only alter the **a** value. So all we need to do is calculate the **a** value.

Substitute the k value into the exponent and solve for **a**.

= 5(^{8}⁄_{25})^{k}

= 5(^{8}⁄_{25})^{1}

= 5(^{8}⁄_{25})

= ^{40}⁄_{25}

= ^{8}⁄_{5}

a = ^{8}⁄_{5} or 1.6

The r = ^{8}⁄_{25} and a = ^{8}⁄_{5} as your final answers.

## General Review[edit]

A quick overview of all the key points you need to remember when it comes to patterns in the series and what each variable represents.

- Although the 'in general' section of the review appears to have S
_{n}subtracted by rS_{n}, you are actually subtracting the first line from the second line --> rS_{n}subtracted by S_{n}. This results in ar^{n}- a instead of what would be misinterpreted as a - ar^{n}

## Applications & Examples[edit]

Here are some examples of a few problems.

*Ex. 1:* Determine S_{12} for the geometric series 8 + 12 + 18 + ...

The first thing you want to recognize is the first term. In this case the first term is 8, a = 8. Then you also want to recognize the number of terms the question has asked you to calculate for, so n = 12.

Now all you are missing is the r value before you can put it into a formula. To calculate for **r**, you want to find the ratio between the first two terms. This requires you to divide the second term by the first term to find out the common ratio.

r = ^{t2}⁄_{t1}

r = ^{12}⁄_{8}

r = ^{3}⁄_{2} or 1.5

Now that you have all **a**, **n**, and **r** you can plug it into this formula and solve for S_{n}

S_{n}= ^{a(1 - (r)n)}⁄_{1-(r)}

= ^{8(1 - (3⁄2)12)}⁄_{1-(3⁄2)}

= ^{8(1 - (531441⁄4096))}⁄_{(-1⁄2)}

= ^{8(-527345⁄4096))}⁄_{(-1⁄2)}

= ^{(-527345⁄512)}⁄_{(-1⁄2)}

= 2059.941406

S_{12} = 2059.94

*Ex. 2:* Now let's do an example where you look for the value of **n**. The sum of 6 + 12 + 24 + ... + t_{n} is 1530. How many terms are in the series?

You are given 2 values, S_{n}= 1530 and a = 6. Determine the **r** value before you can plug it into the formulas to find **n**.

Recall that in order to find the common ratio of the series, you need to first divide the second term of the series by the first and then confirm that pattern with following consecutive terms.

r =^{t2}⁄_{t1}

r = ^{12}⁄_{6}

r = 2

In this case, because we calculated **r** and it resulted in 2, we want to plug it into the formula on the right because r > 1

S_{n} = ^{a((r)n-1)}⁄_{(r)-1}

1530 = ^{6((2)n-1)}⁄_{(2)-1}

1530 = ^{6((2)n-1)}⁄_{(2)-1}

Divide both sides by 6 because the denominator 2 - 1 equals to 1 which leaves the numerator over 1.

255 = ((2)^{n}-1)

Add one to the other side

256 = 2^{n}

and because n is in the exponent, the only way we can solve for n is to flip the entire equation into a logarithm.

n = log_{2}(256)

n = 8

There are a total of 8 terms in the sum.

*Ex. 3:* Here is a question involving geometric series that in applying to a real world setting. A ball is dropped from 8m, it bounces 75% of its height each time. Find the vertical distance travelled after the ball hits the ground for the tenth time.

What we can gather right away is the common ratio, the first term and the number of terms in the series. Immediately, we can calculate the S_{10} by plugging the three values into a formula. **a** = 8, **r** = 0.75, **n** = 10

S_{n}= ^{a(1 - (r)n)}⁄_{1-(r)}

S_{10}= ^{8(1 - (0.75)10)}⁄_{1-(0.75)}

= ^{8(1 - (0.0563135147)}⁄_{1-(0.75)}

= ^{8(0.9436864853)}⁄_{0.25}

= ^{7.549491882}⁄_{0.25}

= 30.1979753

S_{10} = 30.198m

We are not done here because what we were looking for is the total vertical distance travelled, all we've calculated so far is the distance it took for the ball to fall 10 times. Now we need to calculate all the times the ball bounced upwards and add both distances to determine the final distance travelled.

As you can see in the diagram above, we need to calculate the distance travelled when the ball bounced upwards. Since the ball did not start on the ground it has 9 bounces compared to the 10 falls. As well, the very first bounce traveled less than 8m as the height travelled depreciated by 75% after every bounce. To calculate the first term of the upward bounces, you need to multiply 8 by the common ratio, which is 0.75. Here are the new values we can use for the next calculation. **a** = 6m, **r** = 0.75 and **n** = 9

S_{n}= ^{a(1 - (r)n)}⁄_{1-(r)}

S_{9}= ^{6(1 - (0.75)9)}⁄_{1-(0.75)}

= ^{6(1 - (0.0750846863))}⁄_{1-(0.75)}

= ^{6(0.9249153137)}⁄_{0.25}

= ^{5.549491882}⁄_{0.25}

= 22.1979753

S_{9} = 22.198m

Now that we have both vertical distances, we can add them together to get the total vertical distance.

V_{T} = V_{9} + V_{10} where V_{T} indicates the total vertical distance.

= 22.198m + 30.198m

= 52.39593506

V_{T} = 52.396m

## Resources[edit]

Here are some useful links that will help you understand geometric series better. You can also use these links to study.

- https://studywell.com/wp-content/uploads/2015/04/GeometricSeriesExamQuestions.pdf
- https://www.hec.ca/en/cams/help/topics/geometric_sequences_and_series.pdf
- https://docs.google.com/a/deltalearns.ca/viewer?a=v&pid=sites&srcid=ZGVsdGFsZWFybnMuY2F8bXdhbmd8Z3g6NmNiMGQzN2I3ZGM3MWM2Ng
- http://www.cimt.org.uk/projects/mepres/alevel/pure_ch13.pdf
- https://www.mathcentre.ac.uk/resources/uploaded/mc-ty-apgp-2009-1.pdf

Below we have linked concise video that goes over the basics when it comes to geometric series. The person behind the video teaches with short and clear sentences which allows viewers to better understand the concepts.

## References[edit]

11.3 Geometric Series.pdf. docs.google.com/a/deltalearns.ca/viewer?a=v&pid=sites&srcid=ZGVsdGFsZWFybnMuY2F8bXdhbmd8Z3g6NmNiMGQzN2I3ZGM3MWM2Ng.

Boundless. “Boundless Algebra.” Lumen, courses.lumenlearning.com/boundless-algebra/chapter/geometric-sequences-and-series/#:~:text=geometric%20series%3A%20An%20infinite%20sequence%20of%20summed%20numbers%2C%20whose%20terms,be%20seen%20at%20every%20scale.

“Geometric Sequences & Series.” IWrite Math: Pre-Calculus 12 Book, Absolute Value Publications, pp. 323–382.

“Geometric Series Introduction (Video).” Khan Academy, Khan Academy, www.khanacademy.org/math/algebra2/x2ec2f6f830c9fb89:poly-factor/x2ec2f6f830c9fb89:geo-series/v/geo-series-intro.

“Geometric Series.” Varsity Tutors, www.varsitytutors.com/hotmath/hotmath_help/topics/geometric-series#:~:text=To%20find%20the%20sum%20of%20an%20infinite%20geometric%20series%20having,r%20is%20the%20common%20ratio.

Kuang, Yang, and Elleyne Kase. “How to Identify a Term in a Geometric Sequence When You Know Two Nonconsecutive Terms.” Dummies, 18 Apr. 2017, www.dummies.com/education/math/calculus/how-to-identify-a-term-in-a-geometric-sequence-when-you-know-two-nonconsecutive-terms/.

Stapel, Elizabeth. “Arithmetic & Geometric Sequences.” Purplemath, www.purplemath.com/modules/series3.htm.