Geometric Sequences and Geometric Series

Geometric Sequences / Progressions

The terms “sequence” and “progression” are interchangeable. A “geometric sequence” is the same thing as a “geometric progression”. This post uses the term “sequence”… but if you live in a place that tends to use the word “progression” instead, it means exactly the same thing. So, let’s investigate how to create a geometric sequence (also known as a geometric progression).

Pick a number, any number, and write it down. For example:

$5$

Now pick a second number, any number (I’ll choose 3), which we will call the common ratio. Now multiply the first number by the common ratio, then write their product down to the right of the first number:

$5,~15$

Now, continue multiplying each product by the common ratio (3 in my example) and writing the result down… over, and over, and over:

$5,~15,~45,~135,~405,~1,215, ...$

By following this process, you have created a “Geometric Sequence”, a sequence of numbers in which the ratio of every two successive terms is the same.

Vocabulary and Notation

In the example above, 5 is the first term (also called the starting term) of the sequence or progression. To refer to the first term of a sequence in a generic way that applies to any sequence, mathematicians use the notation

$a_1$

This notation is read as “A sub one” and means: the 1st value in the sequence or progression represented by “a”. The one is a “subscript” (value written slightly below the line of text), and indicates the position of the term within the sequence. So $a_1$ represents the value of the first term in the sequence (5 in the example above), and $a_5$ represents the value of the fifth term in the sequence (405 in the example above).

The Common Ratio

Since all of the terms in a Geometric Sequence must be the same multiple of the term that precedes them (3 times the previous term in the example above), this factor is given a formal name (the common ratio) and is often referred to using the variable $R$ (for Ratio). If you multiply any term by this value, you end up with the value of the next term.

For an existing Geometric Sequence, the common ratio can be calculated by dividing any term by its preceding term:

$\dfrac{a_2}{ a_1},~~\text{or}~~\dfrac{a_7}{a_6},~~\text{etc.}$

Every Geometric Sequence has a common ratio between consecutive terms. Examples include:

$1,~2,~4,~8,~16,~...$

$27,~-9,~3~,~-1,~...$

$1,~0.1,~0.01,~0.001,~...$

The common ratio can be positive or negative. It can be a whole number, a fraction, or even an irrational number. No matter what value it has, it will be the ratio of any two consecutive terms in the Geometric Sequence.

Therefore, to test if a sequence of numbers is a Geometric Sequence, calculate the ratio of successive terms in various locations within the sequence. If you calculate the same ratio between any two adjacent terms chosen from the sequence (be sure to put the later term in the numerator, and the earlier term in the denominator), then the sequence is a Geometric Sequence. One of the series shown above can be used to demonstrate this process:

$27,~-9,~3~,~-1,~...$

$\dfrac{-9}{27}=-\dfrac{1}{3}$

$\dfrac{3}{-9}=-\dfrac{1}{3}$

$\dfrac{-1}{3}=-\dfrac{1}{3}$

Since the ratio between adjacent terms was always equal to the same number (negative one third), this is a Geometric Sequence.

Algebraic Description Of A Geometric Sequence

The existence of a common ratio allows us to calculate terms in a generic way:

$a_2~=~a_1 \cdot R$

$a_3~=~a_2 \cdot R$

$~~~~etc.$

Since every line above follows the same pattern, the whole process can be described a bit more generally and compactly by using a variable as the subscript:

$a_n~=~a_{n-1}\cdot R$

This would be read as “A sub N is equal to A sub N minus 1 times the common ratio R”. If $a_n$ refers to the “Nth” term, then $a_{n-1}$ has a subscript that is one less than N, and therefore refers to the term that immediately precedes $a_n$ . A more intuitive way of reading this equation is “Any term may be calculated by multiplying its preceding term by the common ratio”.

These insights allow a complete description of a Geometric Sequence to take a number of forms:

$5,~15,~45,~135, ...$
Specifying the first three or four terms is enough to demonstrate the common ratio

$a_1~=~5, ~~R~=~3$
Specify the first term and the common ratio

$a_1 ~=~5,~~a_n ~=~a_{n-1} \cdot 3$
Specify the first term, with a rule to get you from each term to the next
This is a recursive definition (you must know one or more previous terms)

$a_n~=~5\cdot (3)^{n-1}$
Specify a rule (based on the term number) for calculating the “Nth” term
This is a closed form, or explicit definition (you only need to know the term number)

Note that if a sequence starts with a 5 then grows by a factor 3 from one term to the next, this situation can be modeled using an exponential equation with 5 as its initial value and 3 as its base (with the domain restriction that “n” must be a positive integer). The last equation above uses this exponential model, and provides the fastest way to calculate the Nth term of the sequence. Generalizing this exponential equation approach leads to a description that applies to any Geometric Sequence:

$a_n~=~a_1\cdot R^{(n-1)}$

Why Is (n-1) In The Equation?

If you know the first term of a sequence ( $a_1$ ), how many common ratios do you need to multiply it by to get to the second term of the sequence ( $a_2$ )? Since you seek the very next term, only one use of the common ratio is needed:

$a_2~=~a_1 \cdot R$

How many common ratios are needed to get from the first to the third term?

$a_3~=~a_1 \cdot R\cdot R$

Now generalize the situation based on these two examples. When the term numbers were one apart (2 – 1 = 1), one use of the common ratio was needed to get from one to the other. When the term numbers were two apart (3 – 1 = 2), two uses of the common ratio were needed to get from one to the other. We will always need to use the common ratio as many times as the difference between the two term numbers. For the general case, to get from $a_1$ to $a_n$ , what is the difference between the two term numbers? One less than the value of “n”, or “n – 1”.

Thus the common ratio must be used “n-1” times to get from the value of the first term to the value of the n’th term.

Another Version Of The Equation

While the equation above (and copied below) applies to any geometric sequence

$a_n~=~a_1\cdot R^{(n-1)}$

there is a very similar version you will also see used often, which is the “growth rate” form of the equation:

$a_n~=~a_1\cdot (1+r)^{n-1}$

Note that the only difference between the two is that the “R” in the first equation becomes “1+r” in the second. Why bother with two versions? Because the second is more convenient when working with problems involving exponential growth or decay, or interest rates. Suppose a geometric sequence is describing a 10% growth pattern:

$100, 110, 121, ...$

what is the common ratio, R?

$\dfrac{110}{100}=1.1$

Why does the common ratio end up being 1.1 when the growth rate is 10% or 0.1? The distributive property of multiplication over addition helps explain this:

$~~100(1.1)\\*~\\*=100(1+0.1)\\*~\\*=100+10\\*~\\*=110$

Multiplying 100 by 1 produces the starting value (100). Multiplying 100 by 0.1 produces the amount of growth (10). Adding the two values produces the new total (110).

Note that this is consistent with the idea that multiplying by a number between 0 and 1 (such as 10%) produces a result smaller than the starting value, while multiplying by a number larger than 1 (such as 1.1) will produce a bigger result.

So, when problem gives you a “rate of growth” or “rate of decay”, if you multiply the rate by the starting value, you will only get the amount of change (10 in the example above). This must be added to (growth) or subtracted from (decay) the starting value to produce the ending value.

On the other hand, if you add 1 to the rate of change (expressed as a positive value for growth, or a negative value for decay), you end up with a factor that takes you directly to the ending value in one calculation. In the example above, distributing 100 over (1 + 0.1) produces both the starting value (100) and the amount of growth (10) that must then be added together. However multiplying 100 by the sum of the two numbers in parentheses (1.1) produces a result (110) with starting value and growth already combined.

The “growth rate” form of the equation for a Geometric Sequence is useful whenever a problem is asking about a “rate”. The rate can end up being either positive (growth) or negative (decay). If you encounter a geometric sequence such as:

$100, 80, 64, ...$

where the common ratio (R) is less than 1 (0.8 in this case), don’t let the fact that R is less than one bother you… To fit this into the “growth rate” form, solve the equation

$1+r=R\\*~\\*1+r=0.8\\*~\\*r=-0.2$

so the growth rate in this problem must be a negative 20%, which implies that exponential decay is being modeled, and the “growth rate” form of the equation is:

$a_n~=~100(1-0.2)^{n-1}$

This form of the equation allows you to quickly read the rate of growth or decay, without having to stop and figure out “what decay rate produces a common ratio of 0.8”.

Solving Geometric Sequence Problems

How many possible “unknowns” does either equation for $a_n$ have?

$a_n~=~a_1\cdot R^{n-1}$

$a_n~=~a_1\cdot (1+r)^{n-1}$

Four: $a_n,~a_1,~R~or~r,~and~n$ . Therefore problems involving Geometric Sequences typically ask one of four questions:

What is the value of the Nth term (calculate the value of $a_n$ )?

What is the value of first term (solve for $a_1$ )?

Given a value, what term number must it be (solve for “n”)?

What is the common ratio (solve for “R”) or rate of growth/decay (solve for “r”)?

To answer one of the above questions, you must know (or be given enough information to determine) values for three of the “unknowns” in the equation above. For example, if you are told that $a_{12}=4,096$ , you can conclude that when “n” is 12, $a_n$ is 4,096, so you know two of the three bits of information you would need to answer a question about this sequence. Most Geometric Sequence problems can by solved by:

Determining the values for three of the four unknowns in the equation for $a_n$
Substituting those values into the appropriate equation above
Solving for the only variable remaining

Some problems will be a little more complex, but you should still be able to use the information provided to determine values for three of the four unknowns. Note that problems which ask you to solve for the term number, which is in the exponent, may require the use of logarithms during the solution process.

For example, suppose the only information that a problem provides are values for the 10th and 15th terms. You can find “R” either by a) taking the ratio of the two terms, then taking its 5th root (the number of common ratios needed as factors to get from the 10th to the 15th term), or b) treating the 10th term as $a_1$ , and the 15th term as $a_6$ (their term numbers are 5 apart, just as 10 and 15 are), then using the equation for $a_n$ to find “R”.

Applications of Geometric Sequences in “Real Life”

Geometric Sequences can be thought of as exponential equations with their domains restricted to integers. So they can model situations that involve a constant rate of growth, but where the only inputs that make sense are integers. Examples include:

– Annual size of a population that is growing (or shrinking) at a constant rate

– Value of money in an account that receives periodic fixed rate interest payments

– Maximum height of a bouncing ball after each bounce (when a fixed % of its energy is lost on each bounce)

– Radioactivity levels of a sample over time

Geometric Series

The Nth term of a “series” is the sum of the first N terms of its underlying sequence.

Consider the Geometric Sequence described at the beginning of this post:

$5,~15,~45,~135,~...$

The series based on this sequence is:

$5,~20,~65,~200,~...$

The 3rd term of the Series (65) is the sum of the first three terms of the underlying sequence (5 + 15 + 45), and is typically described using Sigma Notation with the formula for the Nth term of an Geometric Sequence (as derived above):

$S_3=\displaystyle \sum_{i=1}^{3}(5 \cdot 3^{i-1})\\*~\\*=5+15+45\\*~\\*=65$

Formula for the Nth Term

Just as it is sometimes useful to have a formula for the Nth term of an Geometric Sequence, it is also useful to have a formula for the Nth term of an Geometric Series, which allows us to avoid having to a add up a long list of terms.

The fact that Geometric Sequences are based on multiplication creates a useful pattern which leads to the formula we seek. Suppose we wish to find the sum of the first N terms of a Geometric Sequence. Let’s express each term using the formula above, so that it involves only $a_1$ and “R”, and show only the first and last two terms:

$S_n~=~a_1+a_1R^1+~...~+a_1R^{n-2}+a_1R^{n-1}$

Note that in the Nth term, “R” has an exponent of “n-1”, not “n”. If we multiply both sides of the above equation by “R”, we end up with:

$S_n \cdot R~=~a_1\cdot R+a_1R^1 \cdot R~...~+a_1 R^{n-2} \cdot R+a_1R^{n-1} \cdot R\\*~\\*S_n \cdot R~=~a_1R+a_1R^2+~...~+a_1R^{n-1}+a_1R^n$

Now arrange this equation next to the first one, so that we are ready to create a linear combination of the two to solve for $S_n$ , as you have hopefully done when solving systems of linear equations. Lining up like terms above one another, what do you notice?

$S_n~~~~~=~a_1+a_1R+~a_1R^2+~...~+a_1R^{n-2}+a_1R^{n-1}\\*~\\*S_n \cdot R~=~~~~~~~a_1R+a_1R^2+~~...~~+a_1R^{n-2}+a_1R^{n-1}+a_1R^n$

Hmmm… all but two of the like terms on the right side of the equation are exactly the same as one another. So, if we proceed with solving these two equations as a system, and subtract one from the other, then solve for $S_n$ , we get:

$(S_n)-(S_n)(R)~=~a_1-a_1R^n\\*~\\*S_n(1-R)~=~a_1(1-rR^n)$

$S_n~=~\dfrac{a_1(1-R^n)}{(1-R)}~~~~ \text{Sum of a finite Geometric Sequence}$

This formula will produce the Nth term of a Geometric Series, which will equal the sum of the first N terms of its underlying Geometric Sequence.

Infinite Geometric Series

What happens when we take the sum of an infinite number of terms in a sequence?

Let’s explore this by first considering Arithmetic (not Geometric) Sequences. As the number of terms in an Arithmetic Sequence grows, the last term moves farther and farther from zero as the common difference is added repeatedly. So the last term in an Arithmetic Sequence will always approach infinity (either positive infinity, or negative infinity) as the number of terms approaches infinity, except when the common difference is zero.

If an Arithmetic Sequence’s last term approaches infinity, or even if it is constant (due to a common difference of zero) then the sum of all those terms must approach infinity. Therefore, all Arithmetic Series (except 0, 0, 0, …) will approach infinity as N grows large.

Turning to Geometric Sequences, consider what happens to an exponential expression when:
a) the absolute value of “R” is greater than 1
b) the absolute value of “R” is between 0 and 1

When a value greater than 1 is raised to a positive power, it grows… the larger the base, the faster it grows:

$2^3=8\\*~\\*3^3=27\\*~\\*4^3=64$

So, when the common ratio is greater than 1, the Nth term of a Geometric Sequence will also grow towards infinity as N gets larger. Therefore a Geometric Series with an “R” greater than 1 will grow towards infinity with each additional term.

However, when the absolute value of the common ratio is between -1 and 1, a special situation arises. What happens when a value between zero and one is raised to a power? It shrinks… the closer to zero it is, the faster it shrinks towards zero:

$0.9^3=0.729\\*~\\*0.2^3=0.008\\*~\\*0.01^3=0.000001$

Using this insight, and re-examining the formula for the Nth term of a Geometric Sequence:

$a_n~=~a_1\cdot R^{n-1}$

you can conclude that if the value of “R” is between -1 and 1, then the value of Nth term must move closer and closer to zero as N grows. So, the sum of all those terms could perhaps get closer and closer to some limit, which would represent the exact sum of an infinite number of terms.

Now let’s re-examine the formula for the Nth term of a Geometric Series with this in mind:

$S_n~=~\dfrac{a_1(1-R^n)}{(1-R)}$

If the common ratio is a value between -1 and 1, as “n” gets infinitely large $R^n$ must become infinitely close to zero. Substituting zero for $R^n$ in the formula above, and dropping the “n” subscript on the initial “S” because we are considering the sum of all terms, produces:

$S~=~\dfrac{a_1(1-0)}{(1-R)}\\*~\\*S~=~\dfrac{a_1(1)}{(1-R)}$

$S~=~\dfrac{a_1}{(1-R)}~~~~ \text{Sum of an infinite Geometric Sequence when}~|R|<1$

So “S” is the value that the Nth term of the Geometric Series approaches as N becomes infinitely large, which is equal to the sum of all (an infinite number of) terms in the underlying geometric sequence.

An Example

This formula allows us to easily find the sum of the infinite Geometric Sequence

$2,~1,~0.5,~0.25,~...$

by first determining that $a_1=2$ , $R=0.5$ , then using these values in the above formula

$S~=~\dfrac{a_1}{(1-R)}\\*\\S~=~\dfrac{2}{(1-0.5)}\\*\\S~=~\dfrac{2}{0.5}\\*\\S~=~4$

Therefore, the sum of this infinite Geometric Sequence is the integer 4. Did you expect that an infinite sequence of increasingly small fractions would sum to such a round number?

Solving Geometric Series Problems

Just as with Geometric Sequence problems, there are up to four possible unknowns in an Geometric Series problem: $S_n, ~a_1, ~R, ~and ~n$ . If the problem involves an infinite series, there are three unknowns. So the four types of questions that are typically asked are:

What is the of the Nth term of the Series? (Calculate the value of $S_n$ )

What is the value of first term? (Solve for $a_1$ )

What is the common ratio? (Solve for $R$ )

Given a value, what term number must it be? (Solve for “n”… may require using logarithms)

Or, for problems involving infinite Geometric Series:

What is the sum of all terms in the infinite sequence? (Solve for “S”)

To answer one of the above questions, you must know (or be given enough information to determine) values for all but one of the “unknowns” in the equations above. From there, algebra skills should get you to an answer for the question.

Applications of Geometric Series in “Real Life”

An Geometric Sequence describes something that is periodically growing in an exponential fashion (by the same percentage each time), and a Geometric Series describes the sum of those periodic values. Examples of Geometric Series that could be encountered in the “real world” include:

– What is the total number of births over a 20 year period to a population that grows at a fixed percentage each year?

– How much interest will $1,000 invested in a fixed-rate certificate of deposit earn over 30 years?

– How much of a medicine that a patient takes every 8 hours remains in the body after taking it for 48 hours if only 20% remains after each 8 hour period?

– If consumption of a rare material is growing at 5% per year, how much will be consumed in total over the next 10 years?

– Does $0.9\overline{9}~=~1?$

– If a ball that is dropped bounces back to 80% of its previous height after each bounce, how far up and down will it have travelled after 15 bounces?

By Whit Ford

Math tutor since 1992. Former math teacher, product manager, software developer, research analyst, etc.

View all of Whit Ford's posts.

34 comments

mrdardy says:

May 11, 2014 at 6:21 am

Thanks for this thorough examination of the topic of geometric series. I shared this with two of my colleagues this morning.

Reply
1. Whit Ford says:
  
  May 11, 2014 at 8:21 am
  
  Thank you! Suggestions, questions, comments are always welcome.
  
  Reply
  1. mrdardy says:
    
    May 11, 2014 at 8:50 am
    
    One of my favorite problems that tie together this and the arithmetic series post is – Insert two numbers between 4 and 24 so that the first three numbers form a geometric progression while the last three numbers form an arithmetic progression
  2. Whit Ford says:
    
    May 11, 2014 at 8:57 am
    
    Ooooh – I like it! Good algebra problem that feels like it will get a little messy, but tie together problem formulation,sequence, and system skills.
superprime1 says:

August 2, 2014 at 5:16 am

Mr. Whitford may i ask what type of series is this sequence of numbers 1, 7, 19, 37, 61, 91, 127, …if you add the first and second terms youll get 2^3 plus another term 3^3 and so on. Also, are they figurate numbers? Is a study about this before? Thanks you very much.

Reply
1. Whit Ford says:
  
  August 2, 2014 at 7:11 am
  
  There are many, many sequences which are neither arithmetic nor geometric. Some may be able to be described with an explicit (or closed firm) formula, others may only be described with a recursive definition, and some may defy attempts to describe them with a formula.
  
  I regret that your question goes beyond my experience, as I have spent most of my time with arithmetic and geometric sequences, and yours is neither. Perhaps another reader of this page can provide you with an answer.
  
  Reply
Jakob Kanis says:

January 8, 2016 at 6:49 am

Please help me calculating the average growth rate of this series:
1.5 1.6 1.3 1.1 -0.08 3.6 -0.19

Thank you!

Reply
1. Whit Ford says:
  
  January 8, 2016 at 8:01 am
  
  I don’t think I have ever seen a question quite like this… The series is neither arithmetic, nor geometric, but the question asks for an “average growth rate”, not an “average difference”, so I assume we are meant to treat it as though it were a geometric sequence.
  
  Two approaches come to mind:
  
  1) A statistical one: Look at the ratio of every pair of successive terms, and average that ratio. That seems to follow the instructions, but will not necessarily get you from the first term to the last term exactly.
  
  2) A goal oriented one: Divide the last term by the first term, take the 6th root of that ratio, then subtract one to arrive at a “growth rate” (R) that will get you from the first term to the last term after applying it six times: 1.5 * (1+R)^6 = -0.19
  
  Given that I am not really certain about the goal of the question (to calculate the last term exactly, or to describe in an “average” way the change that happens from one term to the next), I am not certain which of the approaches above is being asked for.
  
  Reply
  1. Jakob Kanis says:
    
    January 8, 2016 at 8:23 am
    
    I am a retired missionary, and the numbers are the average quinquennial growth rates of church membership from 1980 – 2015 of a certain region of our denomination. There are two that are negative. I would like to know the average % with which this series changes. Does this help?
  2. Whit Ford says:
    
    January 8, 2016 at 8:56 am
    
    Yes, that helps a lot! I did not realize that the numbers were already growth rates… I thought they were (in effect) the membership numbers.
    
    If the numbers are already growth rates, you can still take the same two approaches… but the calculations will be a little different.
    
    Your average growth rate is (1.5 + 1.6 + 1.3 + 1.1 – 0.08 + 3.6 – 0.19) / 7 = 1.26% This is a number that is easy to calculate and understand. It describes the average growth rate of the congregation over the past 8 periods. However, if you apply that growth rate to the initial membership 7 times, it is not likely to produce the exact final membership number.
    
    If you want the exact mathematical “average” growth rate, you can find it be calculating: [(Membership in in period 8) / (Membership in period 1)] ^ (1/7) then subtract 1 from the result to convert the “periodic growth factor” into a “periodic growth rate” = R. This is the number that will produce an exact result:
    (Membership in period 1)*(1+R)^(7)=(Membership in period 8). Based on the percentages in your comment, it should end up being 1.2547%
  3. Jakob Kanis says:
    
    January 8, 2016 at 9:45 am
    
    Thank you so much. Could you give me the sequence for a scientific calculator so that I can do this? Thanks.
  4. Whit Ford says:
    
    January 8, 2016 at 10:12 am
    
    I showed the steps used to calculate both numbers if you have the membership totals.
    
    The only thing I did not show was calculating the exact 8-period growth rate from the 7 individual ones. To do that
    
    1) convert each percentage into a decimal and add one to it:
    1.5 becomes 1 + 0.015 = 1.015
    -.08 becomes 1 – 0.0008 = 0.9992
    These numbers are the growth factors for each period.
    
    2) multiply all 7 of the values you calculated in step 1 to produce a single number. This number is the 8-period total growth factor. It will be the same number you get by calculating the ratio: (Membership in period 8) / (Membership in period 1)
    
    3) Since there are seven growth periods between the first and last membership period, take the 7th root of the result of step 2. On a scientific calculator, you generally do that by raising the result to the 1/7 power: R ^ (1/7). This number is the overall quinquennial growth factor, which when applied 7 times to the (Membership in period 1), will produce (Membership in period 8).
    
    4) To convert the result of step 3 into a percentage growth rate, subtract one and multiply by 100. This will be the (average) quinquennial percentage growth rate from 1980 to 2015.
    
    5) To calculate an annual percentage growth rate, take the 35th root of the result of step 2 (since there are 5 years in each period, times 7 periods), subtract 1, and multiply by 100.
  5. Jakob Kanis says:
    
    January 8, 2016 at 11:47 am
    
    Your help is very much appreciated. I was able to calculate the average annual growth (geometrical mean) rate with positive numbers but not with negative ones. Best wishes!
  6. Whit Ford says:
    
    January 8, 2016 at 12:12 pm
    
    You are most welcome.
    
    Many people neglect to turn rates of growth or decay into their corresponding factors before using them.
    
    (Starting balance)(growth/decay rate) = (amount of change)
    
    This is not a big deal if only applying them over one period, as the (amount of change) can easily be added to the (Starting balance) to arrive at an (Ending balance):
    
    (Ending balance) = (Starting Balance) + (Starting Balance)(growth/decay rate)
    
    However, when making multi-period calculations, the above approach becomes very time consuming. If you remember some of your high school algebra, you might notice a common factor on the right side above, which can be factored out:
    
    (Ending balance) = (Starting Balance)[1 + (growth/decay rate)]
    
    The amount in square brackets is what I refer to as the “growth factor”, and is the quantity that gets you directly to an ending balance. This allows the growth factors for successive periods to all be (relatively easily) multiplied together to arrive at a single “net” growth factor that describes the growth across all of the periods:
    
    (Starting Balance) * [1 + (period 1 growth/decay rate)] = (Per 1 Ending balance)
    (Per 1 Ending Balance) * [1 + (period 2 growth/decay rate)] = (Per 2 Ending balance)
    
    or
    
    (Starting Balance) * [1 + (period 1 growth/decay rate)] * [1 + (period 2 growth/decay rate)] = (Per 2 Ending balance)
    
    so
    
    [1 + (period 1 growth/decay rate)] * [1 + (period 2 growth/decay rate)] = (net growth factor over Periods 1 and 2)
    
    By taking the appropriate root of the “net” growth factor, then subtracting one and expressing it as a percentage, you arrive at an “average growth/decay rate”.
Helen says:

May 3, 2016 at 6:31 am

Thank you for this article. You touch on using logarithms in the solutions of geometric sequences. I have a question as follows:
Find how many terms there are in this geometric sequence:
-1, 2, -4, 8, …….., -16 777 216

I am trying to use logs to solve it (the question is in a logs exercise) and I end up needing to find the log of -2 (the common ratio), which of course is not possible. I get the correct answer when I use +2 instead, and I have seen elsewhere a comment aluding to the use of the modulus value, but I have no explanation as to why this would be considered acceptable.

Please could you help me by explaining? Many thanks.

Reply
1. Whit Ford says:
  
  May 3, 2016 at 6:51 am
  
  Helen,
  
  This problem can be solved by considering sign and magnitude separately. Since the signs of the first and last terms are negative, there must be an odd number of negative factors in the last term (N must be odd, while the base is -2 as you said). Or to use the term “modulus”, the exponent mod 2 must equal 1 (there must be a remainder of 1 when the exponent is divided by 2).
  
  Now we turn to considering the magnitude, and to do that we ignore the sign of the base (for the reason you mention).
  
  So, after using the base change formula to evaluate the log on a calculator, you will need to verify that your solution is indeed odd. If it is not odd, then it violates the requirement identified in the first paragraph, which would mean that there is no solution to this problem.
  
  Does that answer your question?
  
  Reply
  1. Helen says:
    
    May 3, 2016 at 7:14 am
    
    Thank you for such a quick response, Mr Ford.
    
    I understood what you said, apart from ‘now we turn to considering the magnitude, and to do that we ignore the sign of the base…..’
    
    Why can we do that? In every other bit of maths we cannot ignore a negative sign, so why can we do that here? (I apologise if I am missing something obvious here).
  2. Whit Ford says:
    
    May 3, 2016 at 7:22 am
    
    Well, we are not ignoring the sign of the base entirely, just analyzing sign and magnitude in separate steps. Some problems cannot be solved in a completely sequential fashion.
    
    For example, when solving a quadratic equation in factored form that is equal to zero, we break the problem up into two pieces (using the zero product property), and solve each piece separately. This problem is similar, in that the mathematical tool we wish to use (logarithms) will not work when the base is negative.
    
    By using your understanding of both patterns at work in the problem (sign, magnitude), and analyzing each separately, you can find, and defend, a solution.
Helen says:

May 3, 2016 at 9:07 am

Thank you very much, sir. I will need to have a think about it as my understanding is not up to yours. But I will get there. Your help is much appreciated, thank you.

Reply
Sasha Alexa says:

July 17, 2016 at 11:57 am

Hello, what if the ratio is a percentage but the terms must be bigger than the previous term, for example 2% bigger?

This is a problem I have encountered and I cannot answer because the answer i get is either way too small & far or way too big and also far from the answer.

-A town with a current population of 150, 000 citizens is growing at the rate of 2% per year. Find the population of the town 20 years from now.

Reply
1. Whit Ford says:
  
  July 17, 2016 at 12:11 pm
  
  In your example, they are not giving you the “common ratio”, but instead are giving you the “growth rate”. So, you can use either the $a_1 \cdot (1 + r)^{n-1}$ form of the common ratio, where “r” is the rate of growth (0.02) or the $a_1 \cdot (R)^{n-1}$ form where “R” is the common ratio (1.02 in your example). So, your example describes a geometric sequence where
  r = 0.2 or R = 1.02
  n = 20
  $a_1$ = 150,000
  and they are asking for $a_n$
  
  Reply
  1. Alex says:
    
    July 17, 2016 at 1:02 pm
    
    Thank you for the answer.
    
    If A sub n = 150,000×1.02^19
    A sub n = 218,521.68 (rounded off) is this correct?
    
    But the formula i was taught is a sub 1 × r^n-1 so if i followed that…
    150000×0.2^19
    = 7.86432×10^-09 on the calcu…
    
    By the way, is r & R different?
    
    Thank you very much in advance.
  2. Whit Ford says:
    
    July 17, 2016 at 2:10 pm
    
    Your first calculation is correct, your second is not. R and r are different. As used in my blog post above, but applied to your question, R = 1.02 while r = 0.02. “R” is the “common ratio” of a geometric sequence, while “r” is the growth or decay rate in the problem… which must have a 1 added to it to become the common ratio of a Geometric Sequence.
  3. Alexa says:
    
    July 17, 2016 at 5:36 pm
    
    Ahh, I could see now. I have read your rule about ratios < 1 and the growth rates. So, to conclude what I have learned, you would have to add 1 to the r if r is between 1 & 0 in order for it to grow?
  4. Whit Ford says:
    
    July 17, 2016 at 5:46 pm
    
    If you are given a “growth rate” or a “decay rate”, you must add 1 to it before raising it to the power of the number of time periods.
    
    If you are given a “common ratio”, do NOT add 1 to it before raising it to a power.
    
    If the growth rate is 257%, the common ratio will be 1 + 2.57 = 3.57
    If the rate of decay is 35%, the common ratio will be 1 – 0.35 = 0.65
    
    If the common ratio is 1.25, the growth rate is 1.25 – 1 = 0.25 or 25%
    If the common ratio is 0.78, the growth rate is 0.78 – 1 = -0.22 or -22% (actually decay, not growth)
Rohan says:

March 14, 2017 at 2:22 am

I need help finding the common ratio of a geometric series when I have been given:
– the first term
– the total sum
– the number of terms

Reply
1. Whit Ford says:
  
  March 14, 2017 at 8:45 am
  
  You state that you have been given three values:
  $~~a_1$
  $~~S_n$
  $~~n$
  Given that one of the above is the sum of all the terms, you need to use the formula for a Geometric Series (as you state).
  
  That formula is:
  $S_n~=~\dfrac{a_1(1-R^n)}{(1-R)}$
  
  Once you have substituted the three values you know into the formula, the only remaining unknown will be the common ratio $R$ . Unfortunately, this variable occurs in both the numerator and the denominator, and one of the two instances is raised to the power $n$ … so once you have done some algebra to get $R$ out of the denominator and collect like terms, you end up having to solve an “Nth” degree polynomial:
  
  $R^n~-~\dfrac{S_n}{a_1}R~+~(\dfrac{S_n}{a_1}~-~1)~=~0$
  
  If $n~=~2$ , you would only have to solve a quadratic, but if the series has more than 2 terms, you would probably need to use the Rational Roots Theorem and Synthetic Division to find values of $R$ that solve this equation. You could also use a graphing calculator or a web site like Wolfram Alpha to solve the resulting expression for you numerically.
  
  Reply
angela says:

May 16, 2017 at 12:50 am

help solving those questions please
1) if a starting salary is $28 000, and one expects to receive an annual increase of 6%, algebraically determine the salary at the end of the of the eight years of work.
2) Rick buys a brand new fishing boat for $7000. Each year that he keeps boat, the value depreciates by 15%. What would the value of the fishing boat be after 10 years of owning it.

Reply
1. Whit Ford says:
  
  May 16, 2017 at 8:15 am
  
  To solve your questions,
  
  1) Identify the information the problem has provided you. Label each quantity (whether given as a number or in words) as one of the following (note that you will only find three of these quantities in your problems above… but which three have you been given?):
  $a_1$ The 1st value in the sequence, or beginning value
  $r$ The interest rate
  $n$ The term number, or number of time intervals or periods
  $a_n$ The last value in the sequence, or ending value
  
  2) Substitute the three quantities you have been provided for the appropriate variables in the generic equation for a Geometric Sequence:
  $a_n~=~a_1\cdot (1+r)^{n-1}$
  
  3) You should now have replaced three of the variables above with quantities. At this point, you will either be able to evaluate the expression (if you have been given values for all three variables to the right of the equal sign), or you will need to solve for the one remaining variable using your algebra skills.
  
  Reply
Nothando says:

February 17, 2018 at 5:39 pm

A student in Botany researches the growth of certain plants. She observed that the plant grew 80mm in the first year. In the second year the height increased by 30mm. From the third year onwards the annual growth of the plant is 4/5 of its growth of the previous year. Determine the height the plant grew during the third year, and the maximum height it will reach.

Reply
1. Whit Ford says:
  
  February 17, 2018 at 8:01 pm
  
  Nothando,
  
  Based on the information in the problem, you can calculate that the tree grew:
  80 mm, 30 mm, 24 mm, 19.2 mm, etc.
  The way the problem is phrased implies that the growth from the 1st to the 2nd year does NOT follow the 4/5 rule, and indeed it does not since 30/80 = 3/8.
  
  So, you can set up a formula to predict the growth in year n:
  $g = 30*(4/5)^{n-2}$
  I have used (n-2) as the exponent to ensure that its result matches the values I calculated for years 2, 3, and 4 in the list above.
  
  So, now we come to the part of the problem that requires a little careful thinking. If we were to sum the formula for g(n) above for n = 1 to infinity, we would be including an incorrect value for year 1. Evaluating g(1) produces 37.5 mm, which is 42.5 mm less than the actual value of 80. We’ll return to this thought below, but for now let’s check the next two results produced by g(n):
  
  Evaluating g(2) produces 30 mm, as specified in the problem.
  
  Evaluating g(3) produces 24 mm, as calculated manually at the very beginning. This answers the first question asked in the problem.
  
  And to answer the problem’s final question, we need to evaluate:
  $H_{max}=42.5+\displaystyle\sum_{n=1}^{\infty}30*(4/5)^{n-2}$
  But, what is the value of $a_1$ that we should use in the formula for the sum of an infinite geometric series? This may be easier to answer if we change our formula a bit, so that it includes the correct value for the first year, but then continues from n=2:
  $H_{max}=80+\displaystyle\sum_{n=2}^{\infty}30*(4/5)^{n-2}$
  and now we treat 30 as the start of the geometric series ( $a_1$ in the formula for the sum):
  $H_{max}=80+\dfrac{30}{(1-\frac{4}{5})}\\*~~=80+\dfrac{30}{\frac{1}{5}}\\*~\\*~~=80+30*5\\*~~=230 mm$
  
  Reply
tonyzerrer says:

May 19, 2018 at 3:55 pm

Great article. I’ve been considering the Geometric Series formula for a bit and still can’t wrap my brain around why I need to divide by one less than the common ratio. I’m working on an intuitive understanding of the denominator. I understand the creation of the numerator, but the denominator still eludes me. Would you be able to shed light?

In the numerator, I see the difference between the first term and one beyond the last term. I’m not sure how that helps, but I accept it. Why would I then expect the sum of the progression to be that difference divided by one less than the common ratio?

Reply
1. Whit Ford says:
  
  May 19, 2018 at 5:04 pm
  
  I don’t know that I can provide a convincing intuitive understanding of the formula. I rely on my understanding of how the formula was derived algebraically (as shown just in the section titled “Formula for the Nth Term” just before “Infinite Geometric Series”. In order to get all but two of the N terms to cancel out, we had to subtract $S_n\cdot R$ from the left side, and its equivalent as a series from the right side. This step is the one that led to the left side having two $S_n$ terms, and when $S_n$ was factored out of the two terms, that left behind a factor of $1-R$ , which is what ends up in the denominator of the formula.
  
  As to having an intuitive sense of why the formula for the sum of a finite geometric series is what it is, I find myself considering two cases in an attempt to make sense of the formula:
  
  A) If R>1, then each term in the series is greater than the previous one, and the sum of the series equals the first term in the series times the ratio $\dfrac{1-R^n}{1-R}$ . The $R^n$ term will become very large when R>1, so the numerator will be a large negative number. The denominator will (typically) be a small negative number between -1 and 0, which will cause the fraction to become larger than the numerator, leading us to a sum which is usually many times larger than the first term. This makes sense when the underlying sequence represent exponential growth, and therefore does NOT approach any horizontal asymptote as n grows large… the sum of all the terms will grow to infinity as n goes to inifinity, so the “multiplier” of the first term must become huge.
  
  B) If R<1, then each term in the series is smaller than the previous one, and the sum of the series still equals the first term in the series times the ratio $\dfrac{1-R^n}{1-R}$ . The $R^n$ term will become very small when R<1, so the numerator will be a positive number that is a bit less than 1. The denominator will (typically) be a small positive number between 0 and 1, but closer to 0 than the numerator is. This will cause the fraction to become larger than the numerator, leading us to a sum which is usually a small multiple of the first term. This makes sense when the sequence represents exponential decay, and therefore approaches a horizontal asymptote as n grows large… the sum of all the terms will also approach some limit as n goes to infinity, and that limit is not usually a very large multiple of the first term.
  
  Does that help?
  
  Reply
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